Cartesian product of two compact metric spaces $(X,d_X)$ and $(Y,d_Y)$ is again compact.
I will prove using sequential compactness.
Let $(x_n,y_n)$ be any sequence in $X\times Y$.
because $X$ is compact and $(x_n)\in X$ so that $\exists(x_{n_k})$ subsequence such that $(x_{n_k})\to x_0 \in X$
now, $(y_{n_k})\in Y$, and $Y$ is compact, so $\exists (y_{n_{k_l}})$ subsequence of $(y_{n_k})$ such that
$$ (y_{n_{k_l}}) \to y_0\in Y$$
Now can we say that $$(x_{n_{k_l}},y_{n_{k_l}}) \to (x_0, y_0) \in X\times Y$$?
because if it were to be true then we have produced a convergent subsequence and we're done.
Best Answer
Yes, you can say that because the sub-subsequence $x_{n_{k_l}}$ of $x_{n_k}$ also converges to $x_0$ and a product sequence converges iff both component sequences converge (this is a general property of the product topology).