[proofplan] We reduce to the case $n = 2$ by induction and then prove the two-factor case using the [Tube Lemma](/theorems/960). Given an open cover of $X_1 \times X_2$, we fix a point $x_1 \in X_1$ and use compactness of $X_2$ to cover the slice $\{x_1\} \times X_2$ by finitely many cover elements. The Tube Lemma promotes this to a tube $W_{x_1} \times X_2$, and compactness of $X_1$ extracts a finite collection of such tubes covering all of $X_1 \times X_2$. [/proofplan] [step:Reduce to the two-factor case by induction on $n$] We proceed by induction on $n$. The base case $n = 1$ is the statement that $X_1$ is compact, which holds by hypothesis. For the inductive step, suppose the product of any $n - 1$ compact spaces is compact. The product $X_1 \times \cdots \times X_n$ is homeomorphic to $(X_1 \times \cdots \times X_{n-1}) \times X_n$ via the canonical association map \begin{align*} \varphi: X_1 \times \cdots \times X_n &\to (X_1 \times \cdots \times X_{n-1}) \times X_n \\ (x_1, \ldots, x_n) &\mapsto ((x_1, \ldots, x_{n-1}), x_n), \end{align*} which is a homeomorphism because both the product topology on $X_1 \times \cdots \times X_n$ and the iterated product topology on $(X_1 \times \cdots \times X_{n-1}) \times X_n$ are generated by the same subbasis — namely, inverse images of open sets under the coordinate projections. By the inductive hypothesis, $X_1 \times \cdots \times X_{n-1}$ is compact. It therefore suffices to prove that the product of two compact spaces is compact. [guided] The idea is to avoid working with $n$ factors directly. Instead, we observe that the product of $n$ spaces can be viewed as the product of $n - 1$ spaces with one more factor, reducing the problem to the two-factor case. **Base case.** When $n = 1$, the product $X_1$ is compact by hypothesis. **Inductive step.** Assume the result holds for products of $n - 1$ compact spaces. We claim the $n$-fold product $X_1 \times \cdots \times X_n$ is homeomorphic to the two-fold product $(X_1 \times \cdots \times X_{n-1}) \times X_n$, so that the problem reduces to the case of two factors. Define \begin{align*} \varphi: X_1 \times \cdots \times X_n &\to (X_1 \times \cdots \times X_{n-1}) \times X_n \\ (x_1, \ldots, x_n) &\mapsto ((x_1, \ldots, x_{n-1}), x_n). \end{align*} This map is a bijection with inverse $((\hat{x}), x_n) \mapsto (\hat{x}, x_n)$. To see that $\varphi$ is a homeomorphism, note that both product topologies are generated by the same subbasis: the collection of all sets of the form $\pi_i^{-1}(U_i)$ where $U_i \subset X_i$ is open and $\pi_i$ is the $i$-th coordinate projection. Since $\varphi$ preserves these subbasis elements, it is a homeomorphism. By the inductive hypothesis, $Y := X_1 \times \cdots \times X_{n-1}$ is compact. We are therefore reduced to showing: if $Y$ and $X_n$ are compact, then $Y \times X_n$ is compact. [/guided] [/step] [step:Cover each slice $\{x_1\} \times X_2$ by finitely many open sets from the cover] Let $X_1$ and $X_2$ be compact topological spaces and let $\mathcal{U} = \{U_\alpha\}_{\alpha \in A}$ be an open cover of $X_1 \times X_2$. Fix $x_1 \in X_1$. The slice $\{x_1\} \times X_2$ is compact, since the map $\iota_{x_1}: X_2 \to X_1 \times X_2$ defined by $x_2 \mapsto (x_1, x_2)$ is continuous (as a section of the projection $\pi_2$) and its image is $\{x_1\} \times X_2$, which is therefore compact by [Continuous Image of a Compact Space is Compact](/theorems/305). Since $\mathcal{U}$ covers $X_1 \times X_2$, it covers $\{x_1\} \times X_2$ in particular, and compactness of this slice yields a finite subcover: \begin{align*} \{x_1\} \times X_2 \subset U_{\alpha_1} \cup \cdots \cup U_{\alpha_m}. \end{align*} Set $N_{x_1} := U_{\alpha_1} \cup \cdots \cup U_{\alpha_m}$. This is an open set containing $\{x_1\} \times X_2$. [guided] Fix an arbitrary point $x_1 \in X_1$. We want to show that the vertical slice $\{x_1\} \times X_2$ can be covered by finitely many members of $\mathcal{U}$. Why is $\{x_1\} \times X_2$ compact? Define the inclusion map \begin{align*} \iota_{x_1}: X_2 &\to X_1 \times X_2 \\ x_2 &\mapsto (x_1, x_2). \end{align*} This map is continuous because $\pi_1 \circ \iota_{x_1}$ is the constant map $x_2 \mapsto x_1$ (continuous) and $\pi_2 \circ \iota_{x_1}$ is the identity on $X_2$ (continuous), and a map into a product is continuous if and only if each coordinate projection is continuous, by the [Universal Property of the Product Topology](/theorems/292). Since $X_2$ is compact by hypothesis and $\iota_{x_1}$ is continuous, the image $\{x_1\} \times X_2 = \iota_{x_1}(X_2)$ is compact by [Continuous Image of a Compact Space is Compact](/theorems/305). Since $\mathcal{U}$ covers all of $X_1 \times X_2$, it covers $\{x_1\} \times X_2$ in particular. The compactness of this slice yields finitely many indices $\alpha_1, \ldots, \alpha_m$ (depending on $x_1$) such that \begin{align*} \{x_1\} \times X_2 \subset U_{\alpha_1} \cup \cdots \cup U_{\alpha_m}. \end{align*} We define $N_{x_1} := U_{\alpha_1} \cup \cdots \cup U_{\alpha_m}$, a finite union of open sets, hence open, and containing $\{x_1\} \times X_2$. [/guided] [/step] [step:Apply the Tube Lemma to promote each slice cover to a tube] By the [Tube Lemma](/theorems/960), since $N_{x_1}$ is an open set in $X_1 \times X_2$ containing the slice $\{x_1\} \times X_2$ and $X_2$ is compact, there exists an open set $W_{x_1} \subset X_1$ containing $x_1$ such that \begin{align*} W_{x_1} \times X_2 \subset N_{x_1} = U_{\alpha_1} \cup \cdots \cup U_{\alpha_m}. \end{align*} This associates to each $x_1 \in X_1$ an open neighbourhood $W_{x_1}$ and a finite collection $\{U_{\alpha_1}, \ldots, U_{\alpha_m}\}$ from $\mathcal{U}$ that covers the entire tube $W_{x_1} \times X_2$. [guided] The finite union $N_{x_1} = U_{\alpha_1} \cup \cdots \cup U_{\alpha_m}$ is open and contains the slice $\{x_1\} \times X_2$. The [Tube Lemma](/theorems/960) applies because $X_2$ is compact: it guarantees an open neighbourhood $W_{x_1}$ of $x_1$ in $X_1$ such that the entire tube $W_{x_1} \times X_2$ is contained in $N_{x_1}$. The geometric picture is: instead of merely covering a single vertical line $\{x_1\} \times X_2$, we have "fattened" it to a strip $W_{x_1} \times X_2$ that is still covered by the same finite collection from $\mathcal{U}$. The Tube Lemma is the essential ingredient that makes this fattening possible — without it, the finite cover of the slice need not extend to any neighbourhood in the $X_1$-direction. [/guided] [/step] [step:Extract a finite subcover using compactness of $X_1$] As $x_1$ varies over $X_1$, the collection $\{W_{x_1}\}_{x_1 \in X_1}$ is an open cover of $X_1$. Since $X_1$ is compact, there exist finitely many points $x_1^{(1)}, \ldots, x_1^{(k)} \in X_1$ such that \begin{align*} X_1 = W_{x_1^{(1)}} \cup \cdots \cup W_{x_1^{(k)}}. \end{align*} For each $j \in \{1, \ldots, k\}$, the tube $W_{x_1^{(j)}} \times X_2$ is covered by a finite subcollection $\mathcal{U}_j \subset \mathcal{U}$. Since \begin{align*} X_1 \times X_2 = \bigcup_{j=1}^k \left( W_{x_1^{(j)}} \times X_2 \right) \subset \bigcup_{j=1}^k \bigcup_{U \in \mathcal{U}_j} U, \end{align*} the finite collection $\mathcal{U}_1 \cup \cdots \cup \mathcal{U}_k$ is a finite subcover of $\mathcal{U}$ for $X_1 \times X_2$. This proves that $X_1 \times X_2$ is compact, completing the induction. [guided] We have associated to each point $x_1 \in X_1$ an open neighbourhood $W_{x_1} \subset X_1$. The family $\{W_{x_1}\}_{x_1 \in X_1}$ covers $X_1$, and since $X_1$ is compact, finitely many of these neighbourhoods suffice: \begin{align*} X_1 = W_{x_1^{(1)}} \cup \cdots \cup W_{x_1^{(k)}}. \end{align*} Now we assemble the finite subcover of $X_1 \times X_2$. For each $j \in \{1, \ldots, k\}$, the tube $W_{x_1^{(j)}} \times X_2$ is covered by a finite subcollection $\mathcal{U}_j = \{U_{\alpha_1^{(j)}}, \ldots, U_{\alpha_{m_j}^{(j)}}\}$ from the original cover $\mathcal{U}$. Taking the union over all $j$: \begin{align*} X_1 \times X_2 &= \bigcup_{j=1}^k \left( W_{x_1^{(j)}} \times X_2 \right) \subset \bigcup_{j=1}^k \bigcup_{i=1}^{m_j} U_{\alpha_i^{(j)}}. \end{align*} The right-hand side is a finite union of members of $\mathcal{U}$ — it contains at most $m_1 + \cdots + m_k$ sets, all from the original cover. This is the desired finite subcover. With the two-factor case established, the induction from the first step gives compactness of $X_1 \times \cdots \times X_n$ for all $n \ge 1$. [/guided] [/step]