Proving Tychonoff Theorem using the Wallace Theorem 
Prove the Tychonoff theorem using the Wallace theorem and Kuratowki theorem

Wallace Theorem: If $A_\alpha$ is a compact subset of $X_\alpha$ for all $\alpha \in A$, then, for every open subset $W$ of $\prod X_\alpha$ containing $\prod A_\alpha$, there exist basic open subset $\prod U_\alpha$ of $\prod X_\alpha$ (with $U_\alpha = X_\alpha$ for all $\alpha$ in a finite subset $A_0$ of $A$) such that $\prod A_\alpha \subseteq \prod U_\alpha \subseteq W$
Kuratowski Theorem: $X$ is compact iff the projection map $\pi:X\times Y\to Y$ is closed for all spaces $Y$
I'm unable to connect to use the Wallace Theorem, a theorem involving open sets, with the Kuratowski Theorem, a theorem involving closed maps. I think that we have to use the fact that $\pi_\alpha:A_\alpha\times Y\to Y$ is closed for all $\alpha$, but I'm not sure how. Any help is appreciated!
 A: Let $X = \prod_{\alpha \in A} X_\alpha$ be a product of compact spaces.
Let $Y$ be any space, and $p: X \times Y \to Y$ the projection.
Let $C \subseteq X \times Y$ be closed and suppose that $y \notin p[C]$.
It follows that $X \times \{y\}$ (a product of compact subsets) is contained in the open set $(X \times Y)\setminus C$, so Wallace's theorem implies there are open subsets $U_\alpha \subseteq X_\alpha, (y \in) V \subseteq Y$ (where all but finitely many of them are the whole space they're from) such that
$$X \times \{y\} \subseteq \prod_{\alpha \in A} U_\alpha \times V \subseteq (X \times Y)\setminus C$$
We conclude that $U_\alpha=X_\alpha$ for all $\alpha \in A$ and consequently that $V \cap p[C]  = \emptyset$ and this shows $p[C]$ is closed in $Y$ (as $y \notin p[C]$ was arbitrary). As $C$ too was arbitrary, $p$ is a closed map and so
Kuratowski implies that $X$ is compact (as also $Y$ was arbitrary), and Tychonoff's theorem has been proved.
Of course in the context of Engelking, from whence this exercise stems, this is a bit circular, as he makes have use of Tychonoff to show Wallace's theorem in the first place, but the point he makes is that the theorems are about equally "strong" (like different versions of AC are; one implies the other "easily").
