Since $X$ is a sheaf for the Zariski topology it suffices to show the following. Given a surjective flat morphism of affines $f : T' \to T$ we have: $X(T)$ is the equalizer of the two maps $X(T') \to X(T' \times_T T')$. See Topologies, Lemma \ref{topologies-lemma-sheaf-property-fpqc} (there is a little argument omitted here because the lemma cited is formulated for functors defined on the category of all schemes). \medskip\noindent Let $a, b : T \to X$ be two morphisms such that $a \circ f = b \circ f$. We have to show $a = b$. Consider the fibre product $$ E = X \times_{\Delta_{X/S}, X \times_S X, (a, b)} T. $$ By Spaces, Lemma \ref{spaces-lemma-properties-diagonal} the morphism $\Delta_{X/S}$ is a representable monomorphism. Hence $E \to T$ is a monomorphism of schemes. Our assumption that $a \circ f = b \circ f$ implies that $T' \to T$ factors (uniquely) through $E$. Consider the commutative diagram $$ \xymatrix{ T' \times_T E \ar[r] \ar[d] & E \ar[d] \\ T' \ar[r] \ar@/^5ex/[u] \ar[ru] & T } $$ Since the projection $T' \times_T E \to T'$ is a monomorphism with a section we conclude it is an isomorphism. Hence we conclude that $E \to T$ is an isomorphism by Descent, Lemma \ref{descent-lemma-descending-property-isomorphism}. This means $a = b$ as desired. \medskip\noindent Next, let $c : T' \to X$ be a morphism such that the two compositions $T' \times_T T' \to T' \to X$ are the same. We have to find a morphism $a : T \to X$ whose composition with $T' \to T$ is $c$. Choose an affine scheme $U$ and an \'etale morphism $U \to X$ such that the image of $|U| \to |X|$ contains the image of $|c| : |T'| \to |X|$. This is possible by Lemmas \ref{lemma-topology-points} and \ref{lemma-cover-by-union-affines}, the fact that a finite disjoint union of affines is affine, and the fact that $|T'|$ is quasi-compact (small argument omitted). Since $U \to X$ is separated (Lemma \ref{lemma-separated-cover}), we see that $$ V = U \times_{X, c} T' \longrightarrow T' $$ is a surjective, \'etale, separated morphism of schemes (to see that it is surjective use Lemma \ref{lemma-points-cartesian} and our choice of $U \to X$). The fact that $c \circ \text{pr}_0 = c \circ \text{pr}_1$ means that we obtain a descent datum on $V/T'/T$ (Descent, Definition \ref{descent-definition-descent-datum}) because \begin{align*} V \times_{T'} (T' \times_T T') & = U \times_{X, c \circ \text{pr}_0} (T' \times_T T') \\ & = (T' \times_T T') \times_{c \circ \text{pr}_1, X} U \\ & = (T' \times_T T') \times_{T'} V \end{align*} The morphism $V \to T'$ is ind-quasi-affine by More on Morphisms, Lemma \ref{more-morphisms-lemma-etale-separated-ind-quasi-affine} (because \'etale morphisms are locally quasi-finite, see Morphisms, Lemma \ref{morphisms-lemma-etale-locally-quasi-finite}). By More on Groupoids, Lemma \ref{more-groupoids-lemma-ind-quasi-affine} the descent datum is effective. Say $W \to T$ is a morphism such that there is an isomorphism $\alpha : T' \times_T W \to V$ compatible with the given descent datum on $V$ and the canonical descent datum on $T' \times_T W$. Then $W \to T$ is surjective and \'etale (Descent, Lemmas \ref{descent-lemma-descending-property-surjective} and \ref{descent-lemma-descending-property-etale}). Consider the composition $$ b' : T' \times_T W \longrightarrow V = U \times_{X, c} T' \longrightarrow U $$ The two compositions $b' \circ (\text{pr}_0, 1),  b' \circ (\text{pr}_1, 1) : (T' \times_T T') \times_T W \to T' \times_T W \to U$ agree by our choice of $\alpha$ and the corresponding property of $c$ (computation omitted). Hence $b'$ descends to a morphism $b : W \to U$ by Descent, Lemma \ref{descent-lemma-fpqc-universal-effective-epimorphisms}. The diagram $$ \xymatrix{ T' \times_T W \ar[r] \ar[d] & W \ar[r]_b & U \ar[d] \\ T' \ar[rr]^c &  & X } $$ is commutative. What this means is that we have proved the existence of $a$ \'etale locally on $T$, i.e., we have an $a' : W \to X$. However, since we have proved uniqueness in the first paragraph, we find that this \'etale local solution satisfies the glueing condition, i.e., we have $\text{pr}_0^*a' = \text{pr}_1^*a'$ as elements of $X(W \times_T W)$. Since $X$ is an \'etale sheaf we find a unique $a \in X(T)$ restricting to $a'$ on $W$.