Assume that $E$ is a compact set. Assume that $(x_n)$ be a sequence in $E$ such that $x_n\to x\in X$. In order to prove that $E$ is closed set, we need to prove $x\in E$. In fact, since $(x_n)\subset E\subset X$, there is a sub-sequence $(x_{n_{k}})$ such that $x_{n_k}\to y\in X$ as $k\to\infty$. On the other hand, we also have $x_{n_k}\to x$ as $k\to\infty$. So $x=y$ from the uniquely of the limitations. Since $y\in E$, we deduce $x\in E$.
There's a mistake:
so we obtain an open cover $K^c \cup V$ of $K$, and because $K$ is compact
You don't know that $K$ is compact. You are trying to prove that $K$ is compact. You only know that $K$ is closed and $F$ is compact. In fact, the mere fact that you don't use $F$ anywhere in your proof should be cause for alarm.
Also, you are confusing covers (which are collections of sets) with their coverage (i.e., union of their component sets).
So, when you write that you take a finite subcover $\Omega$ of $K$, you then say that $K\subset \Omega$, which is not true. What is true is that $$K\subset\bigcup_{A\in\Omega} A$$ which is different!
I advise you to restart the proof again, and here's a couple points to start you off:
- You need to prove that $K$ is compact
- That means you need to prove every open cover of $K$ has a finite subcover
- That means you take some cover, $\mathcal V$, such that $K$ is covered by $\mathcal V$ (you were on point until here).
- Now you need to prove there exists a finite subcover of $\mathcal V$.
And a little hint:
- You know $F$ is compact, so any open cover of $F$ has a finite subcover.
- Can you add one more set to $\mathcal V$ such that the resulting cover will be a cover for $F$?
Aftert your edit:
You still have that misstake of cover vs coverage. So, instead of saying that $K^c\cup V$ is an open cover of $F$, you should say that $\{K^c\}\cup \{V_a\}$ is an open cover of $F$.
I am even more concerned that I think you are confused a bit about what an open cover is. When you say
Now, consider the union $K^c \cup V$. Since $K \subset V$, it follows that $X =K^c \cup K \subset K^c \cup V$, meaning that $X = K^c \cup V$, so $F \subset K^c \cup V$ and $X$ is an open subset of itself
I'm thinking "yes, all he said is true, but it's irrelevant".
An open cover is a collection of open sets, not a collection of sets whose union is open. So, you don't need to prove that $$K^c\cup V$$ is open (even though it is), you need to prove that every element of $$\{K^c\}\cup\{V_a\}$$ is open (which is not hard, it's just that if you don't do that, the proof is incorrect).
Best Answer
Basically what this proof does is it shows that given an open cover $U$ of $F$, we can extend $U$ to cover all of $K$ by adding the open set $F^c$ (this is possible since $F$ is closed). Then since this new open cover covers all of $K$, it must have a finite subcover by compactness, call it $U' \subseteq U \cup \left\{F^c \right\}$. Then $U'$ covers $F$ since $F \subseteq K$, and we can toss out $F^c$ to get an open cover of $F$ that is a finite subcover of $U$. Since $U$ was arbitrary, the claim follows.
Note that we didn't use any particular properties of metric spaces here, this actually holds in any compact topological space $K$.