By Sites, Lemma \ref{sites-lemma-point-site-topos} there is a one to one correspondence between points of the site and points of the associated topos, hence it suffices to prove (1). By Sites, Proposition \ref{sites-proposition-point-limits} the functor $u$ has the following properties: (a) $u(S) = \{*\}$, (b) $u(U \times_V W) = u(U) \times_{u(V)} u(W)$, and (c) if $\{U_i \to U\}$ is an \'etale covering, then $\coprod u(U_i) \to u(U)$ is surjective. In particular, if $U' \subset U$ is an open subscheme, then $u(U') \subset u(U)$. Moreover, by Sites, Lemma \ref{sites-lemma-point-site-topos} we can write $u(U) = p^{-1}(h_U^\#)$, in other words $u(U)$ is the stalk of the representable sheaf $h_U$. If $U = V \amalg W$, then we see that $h_U = (h_V \amalg h_W)^\#$ and we get $u(U) = u(V) \amalg u(W)$ since $p^{-1}$ is exact. \medskip\noindent Consider the restriction of $u$ to $S_{Zar}$. By Sites, Examples \ref{sites-example-point-topological} and \ref{sites-example-point-topology} there exists a unique point $s \in S$ such that for $S' \subset S$ open we have $u(S') = \{*\}$ if $s \in S'$ and $u(S') = \emptyset$ if $s \not \in S'$. Note that if $\varphi : U \to S$ is an object of $S_\etale$ then $\varphi(U) \subset S$ is open (see Proposition \ref{proposition-etale-morphisms}) and $\{U \to \varphi(U)\}$ is an \'etale covering. Hence we conclude that $u(U) = \emptyset \Leftrightarrow s \in \varphi(U)$. \medskip\noindent Pick a geometric point $\overline{s} : \overline{s} \to S$ lying over $s$, see Definition \ref{definition-geometric-point} for customary abuse of notation. Suppose that $\varphi : U \to S$ is an object of $S_\etale$ with $U$ affine. Note that $\varphi$ is separated, and that the fibre $U_s$ of $\varphi$ over $s$ is an affine scheme over $\Spec(\kappa(s))$ which is the spectrum of a finite product of finite separable extensions $k_i$ of $\kappa(s)$. Hence we may apply \'Etale Morphisms, Lemma \ref{etale-lemma-etale-etale-local-technical} to get an \'etale neighbourhood $(V, \overline{v})$ of $(S, \overline{s})$ such that $$ U \times_S V = U_1 \amalg \ldots \amalg U_n \amalg W $$ with $U_i \to V$ an isomorphism and $W$ having no point lying over $\overline{v}$. Thus we conclude that $$ u(U) \times u(V) = u(U \times_S V) = u(U_1) \amalg \ldots \amalg u(U_n) \amalg u(W) $$ and of course also $u(U_i) = u(V)$. After shrinking $V$ a bit we can assume that $V$ has exactly one point lying over $s$, and hence $W$ has no point lying over $s$. By the above this then gives $u(W) = \emptyset$. Hence we obtain $$ u(U) \times u(V) = u(U_1) \amalg \ldots \amalg u(U_n) = \coprod\nolimits_{i = 1, \ldots, n} u(V) $$ Note that $u(V) \not = \emptyset$ as $s$ is in the image of $V \to S$. In particular, we see that in this situation $u(U)$ is a finite set with $n$ elements. \medskip\noindent Consider the limit $$ \lim_{(V, \overline{v})} u(V) $$ over the category of \'etale neighbourhoods $(V, \overline{v})$ of $\overline{s}$. It is clear that we get the same value when taking the limit over the subcategory of $(V, \overline{v})$ with $V$ affine. By the previous paragraph (applied with the roles of $V$ and $U$ switched) we see that in this case $u(V)$ is always a finite nonempty set. Moreover, the limit is cofiltered, see Lemma \ref{lemma-cofinal-etale}. Hence by Categories, Section \ref{categories-section-codirected-limits} the limit is nonempty. Pick an element $x$ from this limit. This means we obtain a $x_{V, \overline{v}} \in u(V)$ for every \'etale neighbourhood $(V, \overline{v})$ of $(S, \overline{s})$ such that for every morphism of \'etale neighbourhoods $\varphi : (V', \overline{v}') \to (V, \overline{v})$ we have $u(\varphi)(x_{V', \overline{v}'}) = x_{V, \overline{v}}$. \medskip\noindent We will use the choice of $x$ to construct a functorial bijective map $$ c : |U_{\overline{s}}| \longrightarrow u(U) $$ for $U \in \Ob(S_\etale)$ which will conclude the proof. See Lemma \ref{lemma-stalk-gives-point} and its proof for a description of $|U_{\overline{s}}|$. First we claim that it suffices to construct the map for $U$ affine. We omit the proof of this claim. Assume $U \to S$ in $S_\etale$ with $U$ affine, and let $\overline{u} : \overline{s} \to U$ be an element of $|U_{\overline{s}}|$. Choose a $(V, \overline{v})$ such that $U \times_S V$ decomposes as in the third paragraph of the proof. Then the pair $(\overline{u}, \overline{v})$ gives a geometric point of $U \times_S V$ lying over $\overline{v}$ and determines one of the components $U_i$ of $U \times_S V$. More precisely, there exists a section $\sigma : V \to U \times_S V$ of the projection $\text{pr}_U$ such that $(\overline{u}, \overline{v}) = \sigma \circ \overline{v}$. Set $c(\overline{u}) = u(\text{pr}_U)(u(\sigma)(x_{V, \overline{v}})) \in u(U)$. We have to check this is independent of the choice of $(V, \overline{v})$. By Lemma \ref{lemma-cofinal-etale} the category of \'etale neighbourhoods is cofiltered. Hence it suffice to show that given a morphism of \'etale neighbourhood $\varphi : (V', \overline{v}') \to (V, \overline{v})$ and a choice of a section $\sigma' : V' \to U \times_S V'$ of the projection such that $(\overline{u}, \overline{v'}) = \sigma' \circ \overline{v}'$ we have $u(\sigma')(x_{V', \overline{v}'}) = u(\sigma)(x_{V, \overline{v}})$. Consider the diagram $$ \xymatrix{ V' \ar[d]^{\sigma'} \ar[r]_\varphi & V \ar[d]^\sigma \\ U \times_S V' \ar[r]^{1 \times \varphi} & U \times_S V } $$ Now, it may not be the case that this diagram commutes. The reason is that the schemes $V'$ and $V$ may not be connected, and hence the decompositions used to construct $\sigma'$ and $\sigma$ above may not be unique. But we do know that $\sigma \circ \varphi \circ \overline{v}' = (1 \times \varphi) \circ \sigma' \circ \overline{v}'$ by construction. Hence, since $U \times_S V$ is \'etale over $S$, there exists an open neighbourhood $V'' \subset V'$ of $\overline{v'}$ such that the diagram does commute when restricted to $V''$, see Morphisms, Lemma \ref{morphisms-lemma-value-at-one-point}. This means we may extend the diagram above to $$ \xymatrix{ V'' \ar[r] \ar[d]^{\sigma'|_{V''}} & V' \ar[d]^{\sigma'} \ar[r]_\varphi & V \ar[d]^\sigma \\ U \times_S V'' \ar[r] & U \times_S V' \ar[r]^{1 \times \varphi} & U \times_S V } $$ such that the left square and the outer rectangle commute. Since $u$ is a functor this implies that $x_{V'', \overline{v}'}$ maps to the same element in $u(U \times_S V)$ no matter which route we take through the diagram. On the other hand, it maps to the elements $x_{V', \overline{v}'}$ and $x_{V, \overline{v}}$ in $u(V')$ and $u(V)$. This implies the desired equality $u(\sigma')(x_{V', \overline{v}'}) = u(\sigma)(x_{V, \overline{v}})$. \medskip\noindent In a similar manner one proves that the construction $c : |U_{\overline{s}}| \to u(U)$ is functorial in $U$; details omitted. And finally, by the results of the third paragraph it is clear that the map $c$ is bijective which ends the proof of the lemma.