[proofplan] We reduce the problem to the subbasis level using the [Alexander Subbasis Theorem](/theorems/961). The standard subbasis for the [product topology](/page/Product%20Topology) consists of sets of the form $\pi_\alpha^{-1}(U_\alpha)$, where $U_\alpha \subset X_\alpha$ is open. Given an arbitrary cover of $\prod_{\alpha \in A} X_\alpha$ by such subbasis elements, we derive a contradiction from the assumption that no finite subcover exists. The key mechanism is that the failure of finite subcovers forces, for each index $\alpha$, the existence of a point in $X_\alpha$ not covered by the $\alpha$-indexed subbasis sets — and assembling these points yields an element of the product that escapes the entire cover, contradicting the assumption that the subbasis sets cover the product. [/proofplan] [step:Identify the standard subbasis for the product topology] Let $X = \prod_{\alpha \in A} X_\alpha$ carry the [product topology](/page/Product%20Topology), and for each $\alpha \in A$, let $\pi_\alpha: X \to X_\alpha$ denote the canonical projection. By the definition of the product [topology](/page/Topology), a subbasis $\mathcal{S}$ for $X$ is given by the collection \begin{align*} \mathcal{S} = \bigl\{ \pi_\alpha^{-1}(U_\alpha) : \alpha \in A, \; U_\alpha \in \tau_\alpha \bigr\}, \end{align*} where $\tau_\alpha$ denotes the topology on $X_\alpha$. Every [open set](/page/Open%20Set) in the product topology is a union of finite intersections of members of $\mathcal{S}$. [guided] The product topology on $X = \prod_{\alpha \in A} X_\alpha$ is defined as the coarsest topology making every projection $\pi_\alpha: X \to X_\alpha$ [continuous](/page/Continuity). By the [Universal Property of the Product Topology](/theorems/962), this topology has a natural subbasis consisting of preimages of open sets under projections: \begin{align*} \mathcal{S} = \bigl\{ \pi_\alpha^{-1}(U_\alpha) : \alpha \in A, \; U_\alpha \in \tau_\alpha \bigr\}. \end{align*} A basic open set in the product topology is a finite intersection $\pi_{\alpha_1}^{-1}(U_{\alpha_1}) \cap \cdots \cap \pi_{\alpha_k}^{-1}(U_{\alpha_k})$, and every open set is a union of such finite intersections. This subbasis is the handle by which the Alexander Subbasis Theorem will grip the product topology. The crucial feature of these subbasis elements is that each $\pi_\alpha^{-1}(U_\alpha)$ constrains only the $\alpha$-th coordinate — it places no restriction on any other coordinate. This "one coordinate at a time" structure is what makes the argument work. [/guided] [/step] [step:Apply the Alexander Subbasis Theorem to reduce to subbasis covers] By the [Alexander Subbasis Theorem](/theorems/961), $X$ is [compact](/page/Compact%20Space) if and only if every cover of $X$ by members of $\mathcal{S}$ has a finite subcover. We prove that every such subbasis cover admits a finite subcover. Let $\mathcal{C} \subset \mathcal{S}$ be a cover of $X$ by subbasis elements. For each $\alpha \in A$, define the sub-collection \begin{align*} \mathcal{C}_\alpha = \bigl\{ U_\alpha \in \tau_\alpha : \pi_\alpha^{-1}(U_\alpha) \in \mathcal{C} \bigr\}. \end{align*} Each member of $\mathcal{C}$ belongs to exactly one $\mathcal{C}_\alpha$ (since a subbasis element $\pi_\alpha^{-1}(U_\alpha)$ is indexed by a single $\alpha$), so $\mathcal{C}$ partitions into the families $\{\mathcal{C}_\alpha\}_{\alpha \in A}$. [guided] The [Alexander Subbasis Theorem](/theorems/961) states that a [topological](/page/Topology) space is compact if and only if every cover by subbasis elements (from a fixed subbasis) has a finite subcover. The theorem requires a fixed subbasis — we use $\mathcal{S}$ as defined above. The Alexander Subbasis Theorem therefore reduces the problem of showing $X$ is compact (which requires checking *all* open covers) to checking only covers by members of $\mathcal{S}$. Suppose $\mathcal{C} \subset \mathcal{S}$ covers $X$. Each element of $\mathcal{C}$ has the form $\pi_\alpha^{-1}(U_\alpha)$ for some specific index $\alpha$ and some [open set](/page/Open%20Set) $U_\alpha \in \tau_\alpha$. We sort $\mathcal{C}$ by index: for each $\alpha \in A$, we collect all the open sets $U_\alpha$ whose preimage under $\pi_\alpha$ appears in $\mathcal{C}$: \begin{align*} \mathcal{C}_\alpha = \bigl\{ U_\alpha \in \tau_\alpha : \pi_\alpha^{-1}(U_\alpha) \in \mathcal{C} \bigr\}. \end{align*} This is the family of "constraints in the $\alpha$-th coordinate" imposed by the cover $\mathcal{C}$. [/guided] [/step] [step:Show by contradiction that no $\mathcal{C}_\alpha$ covers $X_\alpha$, and select an uncovered point in each factor] Suppose, for the sake of contradiction, that $\mathcal{C}$ has no finite subcover. We claim that for every $\alpha \in A$, the family $\mathcal{C}_\alpha$ fails to cover $X_\alpha$. To see this, suppose some $\mathcal{C}_{\beta}$ covers $X_\beta$. Since $X_\beta$ is [compact](/page/Compact%20Space) by hypothesis, there exist finitely many sets $U_{\beta,1}, \ldots, U_{\beta,m} \in \mathcal{C}_\beta$ with $X_\beta = U_{\beta,1} \cup \cdots \cup U_{\beta,m}$. Then \begin{align*} X = \pi_\beta^{-1}(X_\beta) = \pi_\beta^{-1}(U_{\beta,1} \cup \cdots \cup U_{\beta,m}) = \pi_\beta^{-1}(U_{\beta,1}) \cup \cdots \cup \pi_\beta^{-1}(U_{\beta,m}), \end{align*} which is a finite subcover of $\mathcal{C}$. This contradicts our assumption. Therefore, for every $\alpha \in A$, the family $\mathcal{C}_\alpha$ does not cover $X_\alpha$, and there exists a point $x_\alpha \in X_\alpha \setminus \bigcup \mathcal{C}_\alpha$. [guided] This is the heart of the argument. We assume for contradiction that no finite subfamily of $\mathcal{C}$ covers $X$, and we extract a consequence for each individual coordinate space. Fix any index $\beta \in A$ and suppose the family $\mathcal{C}_\beta = \{U_\beta \in \tau_\beta : \pi_\beta^{-1}(U_\beta) \in \mathcal{C}\}$ covers $X_\beta$. Then $\{U_\beta\}_{U_\beta \in \mathcal{C}_\beta}$ is an open cover of $X_\beta$. Since $X_\beta$ is compact by hypothesis, there exist finitely many elements $U_{\beta,1}, \ldots, U_{\beta,m} \in \mathcal{C}_\beta$ with \begin{align*} X_\beta = U_{\beta,1} \cup \cdots \cup U_{\beta,m}. \end{align*} Applying $\pi_\beta^{-1}$ to both sides, and using the fact that preimages commute with unions, \begin{align*} X = \pi_\beta^{-1}(X_\beta) = \pi_\beta^{-1}\bigl(U_{\beta,1} \cup \cdots \cup U_{\beta,m}\bigr) = \pi_\beta^{-1}(U_{\beta,1}) \cup \cdots \cup \pi_\beta^{-1}(U_{\beta,m}). \end{align*} The equality $X = \pi_\beta^{-1}(X_\beta)$ holds because every element of $X$ has its $\beta$-th coordinate in $X_\beta$. (If any $X_\alpha = \varnothing$, then $X = \varnothing$ is compact and there is nothing to prove; we may therefore assume all $X_\alpha$ are nonempty.) The right-hand side is a finite union of members of $\mathcal{C}$, giving a finite subcover — contradicting our assumption. Since no $\mathcal{C}_\alpha$ can cover $X_\alpha$, for every $\alpha \in A$ there exists a point \begin{align*} x_\alpha \in X_\alpha \setminus \bigcup \mathcal{C}_\alpha. \end{align*} This point $x_\alpha$ is not contained in any [open set](/page/Open%20Set) from $\mathcal{C}_\alpha$. The Axiom of Choice is used here: we simultaneously select one such point from each $X_\alpha \setminus \bigcup \mathcal{C}_\alpha$, which is a nonempty set for each $\alpha$. (In fact, the full strength of the Axiom of Choice is precisely what is needed — Tychonoff's Theorem is equivalent to the Axiom of Choice in ZF.) [/guided] [/step] [step:Construct a point in $X$ not covered by $\mathcal{C}$, reaching a contradiction] By the Axiom of Choice, define $\hat{x} = (x_\alpha)_{\alpha \in A} \in X$, where each $x_\alpha$ is chosen as above. Since $\mathcal{C}$ covers $X$, there exists some $\pi_\gamma^{-1}(U_\gamma) \in \mathcal{C}$ with $\hat{x} \in \pi_\gamma^{-1}(U_\gamma)$. This means $x_\gamma = \pi_\gamma(\hat{x}) \in U_\gamma$. But $U_\gamma \in \mathcal{C}_\gamma$, and $x_\gamma$ was chosen to satisfy $x_\gamma \notin \bigcup \mathcal{C}_\gamma$, so $x_\gamma \notin U_\gamma$. This is a contradiction. Therefore the assumption that $\mathcal{C}$ has no finite subcover is false. By the [Alexander Subbasis Theorem](/theorems/961), $X = \prod_{\alpha \in A} X_\alpha$ is [compact](/page/Compact%20Space). [guided] We now assemble the point that produces the contradiction. Define $\hat{x} \in X$ by specifying its $\alpha$-th coordinate as the point $x_\alpha$ chosen in the previous step: \begin{align*} \hat{x} = (x_\alpha)_{\alpha \in A} \in \prod_{\alpha \in A} X_\alpha. \end{align*} Since $\mathcal{C}$ is a cover of $X$, the point $\hat{x}$ must belong to some member of $\mathcal{C}$. Every member of $\mathcal{C}$ has the form $\pi_\gamma^{-1}(U_\gamma)$ for some $\gamma \in A$ and $U_\gamma \in \mathcal{C}_\gamma$. So there exists $\gamma \in A$ and $U_\gamma \in \mathcal{C}_\gamma$ with \begin{align*} \hat{x} \in \pi_\gamma^{-1}(U_\gamma). \end{align*} By definition of the preimage, $\pi_\gamma(\hat{x}) \in U_\gamma$, i.e., $x_\gamma \in U_\gamma$. But we chose $x_\gamma \in X_\gamma \setminus \bigcup \mathcal{C}_\gamma$, which means $x_\gamma \notin U_\gamma$ for every $U_\gamma \in \mathcal{C}_\gamma$. In particular, $x_\gamma \notin U_\gamma$. This is a contradiction. The contradiction arose from the assumption that no finite subfamily of $\mathcal{C}$ covers $X$. Therefore every subbasis cover of $X$ has a finite subcover. By the [Alexander Subbasis Theorem](/theorems/961), $X$ is compact. [/guided] [/step]