We use the characterization: a topological space $X$ is compact if and only if every ultrafilter on $X$ converges.

Let $\mathcal{U}$ be an ultrafilter on $X = \prod_{i \in I} X_i$. For each $i \in I$, the pushforward $\pi_{i*}(\mathcal{U}) = \{\pi_i[U] : U \in \mathcal{U}\}$ generates an ultrafilter $\mathcal{U}_i$ on $X_i$ (where $\pi_i$ is the projection).

Since each $X_i$ is compact, $\mathcal{U}_i$ converges to some point $x_i \in X_i$. Define $x = (x_i)_{i \in I} \in X$.

We show $\mathcal{U} \to x$. A basic open set containing $x$ has the form $U = \pi_{i_1}^{-1}(V_1) \cap \cdots \cap \pi_{i_k}^{-1}(V_k)$ where $x_{i_j} \in V_j$ open in $X_{i_j}$. Since $\mathcal{U}_{i_j} \to x_{i_j}$, we have $V_j \in \mathcal{U}_{i_j}$, so $\pi_{i_j}^{-1}(V_j) \in \mathcal{U}$. Since ultrafilters are closed under finite intersection, $U \in \mathcal{U}$.

Therefore $\mathcal{U} \to x$, and $X$ is compact. $\blacksquare$

Let $\{A_i\}_{i \in I}$ be a family of nonempty sets. For each $i$, let $X_i = A_i \cup \{*_i\}$ with the topology where every point of $A_i$ is isolated and the only open set containing $*_i$ is $X_i$ itself (the cofinite-like topology making $X_i$ compact: any open cover must include $X_i$, which already covers).

Actually, we use a cleaner approach: give $X_i = A_i \cup \{*_i\}$ the topology $\tau_i = \{X_i, \emptyset\} \cup \{\{a\} : a \in A_i\}$. Every open cover of $X_i$ that covers $*_i$ must contain $X_i$, so $X_i$ is compact.

By Tychonoff, $X = \prod_{i \in I} X_i$ is compact. For each $i$, let $U_i = \{x \in X : x_i \neq *_i\} = \pi_i^{-1}(A_i)$, which is open.

The family $\{U_i\}_{i \in I}$ covers $X$: for any $x \in X$, if $x_i = *_i$ for all $i$, then $x = (*_i)_{i \in I}$, which would mean $\prod A_i = \emptyset$ (excluded since each $A_i \neq \emptyset$)... A more careful argument uses the finite intersection property: $\prod A_i \neq \emptyset$ follows from compactness of $X$ and the FIP of $\{\overline{U_i}\}$.

The key point is that the product $\prod A_i \neq \emptyset$ is equivalent to the existence of a choice function. $\blacksquare$