We have to show that we can find a morphism $\Spec(A) \to X$ fitting into the diagram. \medskip\noindent Consider the base change $X_A = \Spec(A) \times_Y X$ of $X$. Then $X_A \to \Spec(A)$ is a separated morphism of algebraic spaces (Lemma \ref{lemma-base-change-separated}). Base changing all the morphisms of the diagram above we obtain $$ \xymatrix{ \Spec(K') \ar[r] \ar[d] & \Spec(K) \ar[r] & X_A \ar[d] \\ \Spec(A') \ar[r] \ar[rru] & \Spec(A) \ar@{=}[r] & \Spec(A) } $$ Thus we may replace $X$ by $X_A$, assume that $Y = \Spec(A)$ and that we have a diagram as above. We may and do replace $X$ by a quasi-compact open subspace containing the image of $|\Spec(A')| \to |X|$. \medskip\noindent The morphism $\Spec(A') \to X$ is quasi-compact by Lemma \ref{lemma-quasi-compact-permanence}. Let $Z \subset X$ be the scheme theoretic image of $\Spec(A') \to X$. Then $Z$ is a reduced (Lemma \ref{lemma-scheme-theoretic-image-reduced}), quasi-compact (as a closed subspace of $X$), separated (as a closed subspace of $X$) algebraic space over $A$. Consider the base change $$ \Spec(K') = \Spec(A') \times_{\Spec(A)} \Spec(K) \to X \times_{\Spec(A)} \Spec(K) = X_K $$ of the morphism $\Spec(A') \to X$ by the flat morphism of schemes $\Spec(K) \to \Spec(A)$. By Lemma \ref{lemma-flat-base-change-scheme-theoretic-image} we see that the scheme theoretic image of this morphism is the base change $Z_K$ of $Z$. On the other hand, by assumption (i.e., the commutative diagram above) this morphism factors through a morphism $\Spec(K) \to Z_K$ which is a section to the structure morphism $Z_K \to \Spec(K)$. As $Z_K$ is separated, this section is a closed immersion (Lemma \ref{lemma-section-immersion}). We conclude that $Z_K = \Spec(K)$. \medskip\noindent Let $V \to Z$ be a surjective \'etale morphism with $V$ an affine scheme (Properties of Spaces, Lemma \ref{spaces-properties-lemma-quasi-compact-affine-cover}). Say $V = \Spec(B)$. Then $V \times_Z \Spec(A') = \Spec(C)$ is affine as $Z$ is separated. Note that $B \to C$ is injective as $V$ is the scheme theoretic image of $V \times_Z \Spec(A') \to V$ by Lemma \ref{lemma-quasi-compact-scheme-theoretic-image}. On the other hand, $A' \to C$ is \'etale as corresponds to the base change of $V \to Z$. Since $A'$ is a torsion free $A$-module, the flatness of $A' \to C$ implies $C$ is a torsion free $A$-module, hence $B$ is a torsion free $A$-module. Note that being torsion free as an $A$-module is equivalent to being flat (More on Algebra, Lemma \ref{more-algebra-lemma-valuation-ring-torsion-free-flat}). Next, we write $$ V \times_Z V = \Spec(B') $$ Note that the two ring maps $B \to B'$ are \'etale as $V \to Z$ is \'etale. The canonical surjective map $B \otimes_A B \to B'$ becomes an isomorphism after tensoring with $K$ over $A$ because $Z_K = \Spec(K)$. However, $B \otimes_A B$ is torsion free as an $A$-module by our remarks above. Thus $B' = B \otimes_A B$. It follows that the base change of the ring map $A \to B$ by the faithfully flat ring map $A \to B$ is \'etale (note that $\Spec(B) \to \Spec(A)$ is surjective as $X \to \Spec(A)$ is surjective). Hence $A \to B$ is \'etale (Descent, Lemma \ref{descent-lemma-descending-property-etale}), in other words, $V \to X$ is \'etale. Since we have $V \times_Z V = V \times_{\Spec(A)} V$ we conclude that $Z = \Spec(A)$ as algebraic spaces (for example by Spaces, Lemma \ref{spaces-lemma-space-presentation}) and the proof is complete.