Argument for infinite intersection of compacts is compact

Let $K_n$ be compact sets. Is their infinite intersection compact?

My argument is very simple compared to other arguments and I feel like I'm missing something. It goes like this: If the intersection is empty, then it is compact. If it is nonempty, then let $(x_n)$ be a sequence in the intersection. $(x_n) \in K_1$ and since $K_1$ is compact, there exists a convergent sub-sequence $(x_{n_k})$. This sub-sequence is, of course, contained in the intersection, which completes the proof.

Where am I wrong?

• The elements of the subsequence are in the intersection, but you need to know that the limit is also in the intersection. (It is, because compact implies closed, but you need to say that.) – mathguy Nov 16 '17 at 2:03
• @mathguy . Compact implies closed if the space is Hausdorff . But it does not hold in every space. – DanielWainfleet Nov 16 '17 at 4:14
• @DanielWainfleet - What level of topology do you suspect the OP is at? I assume they only do metric spaces or so. – mathguy Nov 16 '17 at 5:07
• @mathguy - You're right in that assumption. Sorry for not pointing that out. – Jalil Compaoré Nov 16 '17 at 7:11

You are right, you are definitely missing something. The limit of the convergent subsequence lies in $K_1$ but you provide no reason it should lie in the intersection.
Let $K = \bigcap_i K_i$. Clearly $K \subseteq K_1$. Since each $K_i$ is closed, so is $K$. Hence $K$ is a closed set inside a compact set $K_1$, thus $K$ is compact.