Proof. Let $C$ be a non-empty collection of non-empty sets. Let $Y$ be the generalized cartesian product of all the elements in $C$ Our objective is show that $Y$ is non-empty.

First, some notations: for each $A\in C$ set $X_A:=A\cup \lbrace A\rbrace$ $D:=\lbrace X_A\mid A\in C\rbrace$ $X$ the generalized cartesian product of all the $X_A$ s, and $p_A$ the projection from $X$ onto $X_A$

We break down the proof into several steps:

1. $Y$ is equipollent to $Z:=\bigcap \lbrace p_A^{-1}(A)\mid A\in C\rbrace$

   An element of $X$ is a function $f: D\to \bigcup D$ such that $f(X_A)\in X_A$ for each $A\in C$ In other words, either $f(X_A)\in A$ or $f(X_A)=A$ An element of $Y$ is a function $g:C\to \bigcup C$ such that $g(A)\in A$ for each $A\in C$ Finally, $h\in p_A^{-1}(A)$ iff $h(X_A)\in A$

   Given $g\in Y$ define $g^*\in X$ by $g^*(X_A):=g(A)\in A$ Since $A$ is arbitrary, $g^*\in Z$ Conversely, given $h\in Z$ define $h'\in Y$ by $h'(A):=h(X_A)$ which is well-defined, since $h(X_A)\in A$ Now, it is easy to see that the function $\phi:Y\to Z$ given by $\phi(g)=g^*$ is a bijection, whose inverse $\phi^{-1}:Z\to Y$ is given by $\phi^{-1}(h)=h'$ This shows that $Y$ and $Z$ are equipollent.

2. Next, we topologize each $X_A$ in such a way that $X_A$ is compact.

   Let $\mathcal{T}_A$ be the coarsest topology containing the cofinite topology on $X_A$ and the singleton $\lbrace A\rbrace$ A typical open set of $X_A$ is either the empty set, or has the form $S\cup \lbrace A\rbrace$ where $S$ is cofinite in $A$

   To show that $X_A$ is compact under $\mathcal{T}_A$ let $\mathcal{D}$ be an open cover for $X_A$ We want to show that there is a finite subset of $\mathcal{D}$ covering $X_A$ If $X_A\in \mathcal{D}$ then we are done. Otherwise, pick a non-empty element $S\cup \lbrace A\rbrace$ in $\mathcal{D}$ so that $A-S\ne \varnothing$ and is finite. By assumption, each element in $A-S$ belongs to some open set in $\mathcal{D}$ So to cover $A-S$ only a finite number of open sets in $\mathcal{D}$ are needed. These open sets, together with $S\cup\lbrace A\rbrace$ cover $X_A$ Hence $X_A$ is compact.

3. Finally, we prove that $Z$ and therefore $Y$ is non-empty.

   Apply Tychonoff's theorem, $X$ is compact under the product topology. Furthermore, $\pi_A$ is continuous for each $A\in C$ Since $\lbrace A\rbrace$ is open in $X_A$ and $A=X_A-\lbrace A\rbrace$ $A$ is closed in $X_A$ and thus so is $p_A^{-1}(A)$ closed in $X$

   To show that $Z$ is non-empty, we employ a characterization of compact space: $X$ is compact iff every collection of closed sets in $X$ having FIP has non-empty intersection. Let us look at the collection $\mathcal{S}:=\lbrace p_A^{-1}(A)\mid A\in C\rbrace$ Given $A_1,\ldots, A_n\in C$ pick an element $a_i\in A_i$ since $A_i\ne \varnothing$ by assumption. Note that this is possible, since there are only a finite number of sets. Define $f:D\to \bigcup D$ as follows:

   
   Since $f(X_{A_i})=a_i \in A_i$ $f\in p_{A_i}^{-1}(A_i)$ for each $i=1,\ldots, n$ Therefore, $$f\in p_{A_1}^{-1}(A_1) \cap \cdots \cap p_{A_n}^{-1}(A_n).$$ Since $p_{A_1}^{-1}(A_1), \ldots, p_{A_n}^{-1}(A_n)$ are arbitrarily picked from $\mathcal{S}$ the collection $\mathcal{S}$ has finite intersection property, and since $X$ is compact, $Z=\bigcap S$ must be non-empty.

This completes the proof.