# Proving that a product of two compact metric spaces is compact. (basic real analysis)

I'm not really sure if this sort of question is frowned upon here, but I don't understand a proposed solution to an exercise I was given. Let me know if it's an inappropriate question and I won't ask similar ones again.

Let K be a compact metric space with metric d.

Exercise: Show that the product $K \times K$ with the metric $d_{MH}((x_1,x_2)(y_1,y_2)) = d(x_1,y_1) + d(x_2,y_2)$ is compact.

Solution (or at least the part of it I'm not understanding):

"Now if we let $(z_n) = (x_n,y _n)$ be a sequence in $K\times K$. Then $x_n$ is a sequence in K, so has a convergent subsequence, say ${x_n}_k$, which converges to some $x\in K$. The corresponding subsequence ${y_n}_k$ is again a sequence in K, so has a convergent subsequence, say ${{y_n}_k}_l$, which converges to $y \in K$. Then ${{x_n}_k}_l$ is a subsequence of the convergent sequence ${x_n}_k$, so it converges to the same element x."

Question: Why do we need the sub subsequences: ${{x_n}_k}_l$ and ${{y_n}_k}_l$? I thought it was sufficient to show that all sequences had a convergent subsequence in order to show that $K\times K$ was compact.

We cannot use the subsequence $(x_{nk}, y_{nk})$ of $(x_n, y_n)$ because although the subsequence $x_{nk}$ of $x_n$ is convergent, there is no guarantee that the corresponding $y_{nk}$ of $y_n$ is convergent. But $y_{nk}$ must have a convergent subsequence and its corresponding subsequence in $x_{nk}$ must also be convergent (being a subsequence of a convergent sequence). This gives us a convergent subsequence of $(x_{nk}, y_{nk})$ and hence a convergent subsequence of $(x_n, y_n)$.