To prove part (1) choose a surjective \'etale morphism $V \to Y$ where $V$ is a scheme. For each $v \in V$ choose an algebraic closure $\kappa(v) \subset k_v$. Consider the morphisms $\overline{x} : \Spec(k_v) \to V \to Y$. By construction of $|Y|$ these cover $|Y|$. \medskip\noindent To prove part (2) we will use the following two facts whose proofs we omit: (\romannumeral1) If $K$ is a field and $\overline{K}$ is algebraic closure then $|\overline{K}| \leq \max\{\aleph_0, |K|\}$. (\romannumeral2) For any algebraically closed field $k$ and any cardinal $\aleph$, $\aleph \geq |k|$ there exists an extension of algebraically closed fields $k \subset k'$ with $|k'| = \aleph$. Now we set $\aleph = \max\{\lambda(X), \lambda(Y)\}^+$. Here $\lambda^+ > \lambda$ indicates the next bigger cardinal, see Sets, Section \ref{sets-section-cardinals}. Now (\romannumeral1) implies that the fields $k_u$ constructed in the first paragraph of the proof all have cardinality bounded by $\lambda(X)$. Hence by (\romannumeral2) we can find extensions $k_u \subset k'_u$ such that $|k'_u| = \aleph$. The morphisms $\overline{x}' : \Spec(k'_u) \to X$ cover $|X|$ as desired. To really finish the proof of (2) we need to show that the schemes $\Spec(k'_u)$ are (isomorphic to) objects of $\Sch_{fppf}$ because our conventions are that all schemes are objects of $\Sch_{fppf}$; the rest of this paragraph should be skipped by anyone who is not interested in set theoretical considerations. By construction there exists an object $T$ of $\Sch_{fppf}$ such that $\lambda(X)$ and $\lambda(Y)$ are bounded by $\text{size}(T)$. By our construction of the category $\Sch_{fppf}$ in Topologies, Definitions \ref{topologies-definition-big-fppf-site} as the category $\Sch_\alpha$ constructed in Sets, Lemma \ref{sets-lemma-construct-category} we see that any scheme whose size is $\leq \text{size}(T)^+$ is isomorphic to an object of $\Sch_{fppf}$. See the expression for the function $Bound$ in Sets, Equation (\ref{sets-equation-bound}). Since $\aleph \leq \text{size}(T)^+$ we conclude. \medskip\noindent The notation $X_s$ in part (3) means the fibre product $\Spec(\kappa(s)) \times_S X$, where $s \in S$ is the point corresponding to $\overline{s}$. Hence part (2) follows from (4) with $Y = \Spec(\kappa(s))$. \medskip\noindent Let us prove (4). Let $X \to Y$ be a morphism of algebraic spaces over $S$. Let $k$ be an algebraically closed field over $S$ of cardinality $> \lambda(X)$. Let $\overline{y} \in Y(k)$ and $x \in |X|$ which map to the same element $y$ of $|Y|$. We have to find $\overline{x} \in X(k)$ mapping to $x$ and $\overline{y}$. Choose a commutative diagram $$ \xymatrix{ U \ar[d] \ar[r] & V \ar[d] \\ X \ar[r] & Y } $$ with $U$, $V$ schemes over $S$ and vertical arrows surjective and \'etale, see Spaces, Lemma \ref{spaces-lemma-lift-morphism-presentations}. Choose a $u \in |U|$ which maps to $x$, and denote $v \in |V|$ the image. We will think of $u = \Spec(\kappa(u))$ and $v = \Spec(\kappa(v))$ as schemes. Note that $V \times_Y \Spec(k)$ is a scheme \'etale over $k$. Hence it is a disjoint union of spectra of finite separable extensions of $k$, see Morphisms, Lemma \ref{morphisms-lemma-etale-over-field}. As $v$ maps to $y$ we see that $v \times_Y \Spec(k)$ is a nonempty scheme. As $v \to V$ is a monomorphism, we see that $v \times_Y \Spec(k) \to V \times_Y \Spec(k)$ is a monomorphism. Hence $v \times_Y \Spec(k)$ is a disjoint union of spectra of finite separable extensions of $k$, by Schemes, Lemma \ref{schemes-lemma-mono-towards-spec-field}. We conclude that the morphism $v \times_Y \Spec(k) \to \Spec(k)$ has a section, i.e., we can find a morphism $\overline{v} : \Spec(k) \to V$ lying over $v$ and over $\overline{y}$. Finally we consider the scheme $$ u \times_{V, \overline{v}} \Spec(k) = \Spec(\kappa(u) \otimes_{\kappa(v)} k) $$ where $\kappa(v) \to k$ is the field map defining the morphism $\overline{v}$. Since the cardinality of $k$ is larger than the cardinality of $\kappa(u)$ by assumption we may apply Algebra, Lemma \ref{algebra-lemma-base-change-Jacobson} to see that any maximal ideal $\mathfrak m \subset \kappa(u) \otimes_{\kappa(v)} k$ has a residue field which is algebraic over $k$ and hence equal to $k$. Such a maximal ideal will hence produce a morphism $\overline{u} : \Spec(k) \to U$ lying over $u$ and mapping to $\overline{v}$. The composition $\Spec(k) \to U \to X$ will be the desired geometric point $\overline{x} \in X(k)$. This concludes the proof of part (4). \medskip\noindent Part (5) is a formal consequence of parts (2) and (4) and Properties of Spaces, Lemma \ref{spaces-properties-lemma-characterize-surjective}.