For each $y \in Y$, the point $(x_0, y) \in N$. Since $N$ is open, there exist open sets $U_y \subseteq X$ and $V_y \subseteq Y$ with $(x_0, y) \in U_y \times V_y \subseteq N$.

The collection $\{V_y : y \in Y\}$ is an open cover of $Y$. Since $Y$ is compact, there is a finite subcover $\{V_{y_1}, \ldots, V_{y_n}\}$.

Set $W = U_{y_1} \cap \cdots \cap U_{y_n}$. Then $W$ is open and $x_0 \in W$ (since $x_0 \in U_{y_i}$ for each $i$).

For any $(x, y) \in W \times Y$: since $\{V_{y_i}\}$ covers $Y$, we have $y \in V_{y_k}$ for some $k$. Then $x \in W \subseteq U_{y_k}$, so $(x, y) \in U_{y_k} \times V_{y_k} \subseteq N$. Therefore $W \times Y \subseteq N$.

Let $\mathcal{O} = \{O_\alpha\}_{\alpha \in A}$ be an open cover of $X \times Y$. We construct a finite subcover.

Step 1: For each $x \in X$, find a tube.

Fix $x \in X$. The slice $\{x\} \times Y$ is homeomorphic to $Y$ (via the map $y \mapsto (x, y)$), hence compact. Since $\mathcal{O}$ covers $\{x\} \times Y$, there exist finitely many $O_{\alpha_1(x)}, \ldots, O_{\alpha_{n(x)}(x)}$ covering $\{x\} \times Y$.

Set $N_x = O_{\alpha_1(x)} \cup \cdots \cup O_{\alpha_{n(x)}(x)}$. Then $N_x$ is open and $\{x\} \times Y \subseteq N_x$.

By the Tube Lemma, there exists an open set $W_x \subseteq X$ with $x \in W_x$ and $W_x \times Y \subseteq N_x$.

Step 2: Extract a finite subcover of $X$.

The collection $\{W_x : x \in X\}$ is an open cover of $X$. Since $X$ is compact, there is a finite subcover $\{W_{x_1}, \ldots, W_{x_m}\}$.

Step 3: Combine.

For each $x_j$, we have $W_{x_j} \times Y \subseteq N_{x_j} = O_{\alpha_1(x_j)} \cup \cdots \cup O_{\alpha_{n(x_j)}(x_j)}$.

Since $X \times Y = \bigcup_{j=1}^m (W_{x_j} \times Y) \subseteq \bigcup_{j=1}^m N_{x_j}$, the collection $\{O_{\alpha_k(x_j)} : 1 \leq j \leq m, \; 1 \leq k \leq n(x_j)\}$ is a finite subcover of $\mathcal{O}$. (There are at most $\sum_{j=1}^m n(x_j)$ sets in this subcover.)

Therefore $X \times Y$ is compact.

By induction on $n$. The base case $n = 1$ is trivial.

For the inductive step, suppose $X_1 \times \cdots \times X_{n-1}$ is compact. There is a natural homeomorphism: $(X_1 \times \cdots \times X_{n-1}) \times X_n \cong X_1 \times \cdots \times X_n$ (using associativity of the product topology). Since $X_1 \times \cdots \times X_{n-1}$ is compact by the inductive hypothesis and $X_n$ is compact, Theorem 3.13 gives that their product is compact.