I am trying to show Every compact metrizable space is second countable
My Attempt:
Let $(X,\mathfrak{T})$ be a compact metrizable space. We wish to show that it has a countable basis.
Then given $C \subset \mathcal{P}(X)$ an open cover of $X$, it has a finite subcover $\{C_n| n \in F, F \text{ is a finite set in } X\}$.
Since $X$ is metrizable, therefore $\forall x \in C_n, \exists B_n \in \mathfrak{B}$ such that $x \in B_n \subset C_n$, where $B_n$ is a metric ball, then $\{B_{n_x}| x \in C_n\}$ is an open cover of $C_n$.
Since the closure $\overline C_n$ is closed, a closed set in a compact space is compact, therefore $\overline C_n$ is compact and covered by finite many $B_{n_x}$. Since $C_n \subset \overline C_n$, so $C_n$ is covered by finite many $B_{n_x}$ as well.
Then $\bigcup\limits_{x \in C_n, n \in F} \{B_{n_x}\}$ is a finite union of finitely many metric balls that covers $X$. Therefore $\{B_{n_x}| x \in C_n, n \in F\}$ is a countable basis.
Can someone check my work? I feel I overkilled it with the closure. Is there a much easier way?
This is not correct. You have produced a countable collection of metric balls that cover $X$, but such a collection need not be a basis. For instance, $X=[0,1]$ is covered by the single metric ball $B_2(0)$, but $\{B_2(0)\}$ is certainly not a basis (the topology it generates is the indiscrete topology!).
A sure-fire sign that what you're doing isn't right is that you haven't actually used any specific open cover of $X$. So for instance, you might just have $C=\{X\}$. In that case, compactness doesn't tell you anything (this cover has a finite subcover for any space), so your argument would work without compactness if it worked at all.
So to find a correct proof, you will need to choose some specific special open covers of $X$ to take finite subcovers of. And to get a basis, you probably want your sets to be "arbitrarily small". This suggests you might try looking at open covers by sets of small diameter.