We prove the contrapositive: if $X$ is not compact, then some subbasis cover has no finite subcover.

Suppose $X$ is not compact. Let $\Sigma$ be the collection of all open covers of $X$ that have no finite subcover. By hypothesis, $\Sigma$ is nonempty.

Step 1: Zorn's lemma. We partially order $\Sigma$ by inclusion (of collections of open sets). If $\{C_\beta\}_{\beta}$ is a chain in $\Sigma$, then $C = \bigcup_\beta C_\beta$ is an open cover with no finite subcover (any finite subcover of $C$ would be a finite subcollection of some $C_\beta$, giving a finite subcover of $C_\beta$, contradicting $C_\beta \in \Sigma$). So $C \in \Sigma$, and Zorn's lemma gives a maximal element $\mathcal{M} \in \Sigma$.

Step 2: Properties of $\mathcal{M}$. The maximal cover $\mathcal{M}$ has no finite subcover, but adding any open set $U \notin \mathcal{M}$ to $\mathcal{M}$ creates a cover with a finite subcover (by maximality). That is, $\mathcal{M} \cup \{U\}$ has a finite subcover $\{U, V_1, \ldots, V_k\}$ with $V_i \in \mathcal{M}$.

This means $X \setminus U \subseteq V_1 \cup \cdots \cup V_k$, or equivalently: $\{V_1, \ldots, V_k\}$ covers $X \setminus U$.

Step 3: Subbasis elements in $\mathcal{M}$. We claim that $\mathcal{M} \cap \mathcal{S}$ does not cover $X$.

Suppose $\mathcal{M} \cap \mathcal{S}$ covers $X$. Every $V \in \mathcal{M}$ is a union of finite intersections of elements of $\mathcal{S}$: $V = \bigcup_\gamma (S_{\gamma,1} \cap \cdots \cap S_{\gamma,n_\gamma})$.

Since $\mathcal{M}$ covers $X$, for each $x \in X$, there exists $V \in \mathcal{M}$ with $x \in V$, hence $x \in S_{\gamma,1} \cap \cdots \cap S_{\gamma, n_\gamma}$ for some $\gamma$. So the sets $S_{\gamma, i}$ that appear cover $X$.

Now, pick any $S \in \mathcal{S}$ that appears in the decomposition of some $V \in \mathcal{M}$. If $S \notin \mathcal{M}$, then by Step 2, there exist $V_1, \ldots, V_k \in \mathcal{M}$ covering $X \setminus S$. But $S$ was needed with other subbasis elements to compose $V$; we can replace $S$ with $\{V_1, \ldots, V_k\}$. Iterating this process (finitely many times for each intersection), we eventually express $X$ as a finite subcover of $\mathcal{M}$, contradicting $\mathcal{M} \in \Sigma$.

More precisely: take any $V \in \mathcal{M}$ and $x \in V$. Then $x \in S_1 \cap \cdots \cap S_m \subseteq V$ for some $S_1, \ldots, S_m \in \mathcal{S}$. If all $S_i \in \mathcal{M}$, we are fine. If some $S_j \notin \mathcal{M}$, by Step 2, finitely many $W_1, \ldots, W_\ell \in \mathcal{M}$ cover $X \setminus S_j$. Then $X = S_j \cup W_1 \cup \cdots \cup W_\ell \supseteq S_j \cup (X \setminus S_j) = X$. But then... we need $S_j$ to be covered, which it is (by elements of $\mathcal{M}$). This gives a finite subcover of $\mathcal{M}$, contradiction.

Therefore $\mathcal{M} \cap \mathcal{S}$ does not cover $X$, and the subbasis cover $\mathcal{S}' \subseteq \mathcal{S}$ formed by taking all subbasis elements that participate in $\mathcal{M}$ fails to cover. Since $\mathcal{M}$ does cover $X$, some element of $\mathcal{M}$ is not a subbasis element. The subbasis elements alone from $\mathcal{M}$ do not suffice to cover, confirming our contrapositive.

The product topology on $P = \prod_\alpha X_\alpha$ has subbasis $\mathcal{S} = \{\pi_\alpha^{-1}(U_\alpha) : \alpha \in A, U_\alpha \text{ open in } X_\alpha\}$. By the Alexander subbasis theorem, it suffices to show every cover of $P$ by subbasis elements has a finite subcover.

Let $\{\pi_{\alpha_i}^{-1}(U_i)\}_{i \in I}$ be a cover of $P$ by subbasis elements. For each $\alpha$, let $\mathcal{U}_\alpha = \{U_i : \alpha_i = \alpha\}$. If $\mathcal{U}_\alpha$ covers $X_\alpha$ for some $\alpha$, then by compactness of $X_\alpha$, finitely many suffice, giving a finite subcover of $P$.

If no $\mathcal{U}_\alpha$ covers $X_\alpha$, then for each $\alpha$, there exists $x_\alpha \in X_\alpha \setminus \bigcup \mathcal{U}_\alpha$. Then the point $(x_\alpha) \in P$ is not in any $\pi_\alpha^{-1}(U_i)$ (since $x_{\alpha_i} \notin U_i$ for each $i$). But $\{\pi_{\alpha_i}^{-1}(U_i)\}$ is supposed to cover $P$. Contradiction.

(Note: the step "for each $\alpha$, there exists $x_\alpha$" uses the axiom of choice.)