Hi this is a problem from Rudin's Princ. of Mathematics. I was hoping someone could check this part of my proof for the following question, comments would be very appreciated!:
$25.$ Prove that every compact metric space $K$ has a countable base, and that $K$ is therefor separable.
Consider the neighborhoods of radii $\frac{1}{n}$ for every $k_{i} \in K$. Then we have that $K \subset \bigcup\limits_{i=1}^{\infty} N_{1/n}(k_{i})$, and so $\bigcup\limits_{i=1}^{\infty} N_{1/n}(k_{i})$ is an open cover of $K$. Since $K$ is compact, there is a finite sub-collection of this set, let the sequence of these sets be $\{V_{\alpha}\}$. Then we claim that this is a countable base for $K$. Since it is finite, it is clearly countable. Then for any $k \in K$ and open set $G \subset K$ such that $k \in G$, we have that $k \in V_{\alpha}$ for some $\alpha$. Then since these are open sets, there is a neighborhood with a sufficiently small radius that is a subset of an interior neighborhood of $G$.
Ie. since $G$ is open then there is a neighborhood $N_{r}(k) \subset G$. Then let $\delta < \frac{r}{2}$. Since $k \in V_{\alpha}$, we have that there is a neighborhood $N_{\delta}(k)$ for small enough $\delta$ such that $N_{\delta}(k) \in V_{\alpha}$, since $V_{\alpha}$ is open such a neighborhood exists. Then $N_{\delta}(k) \subset N_{r}(k) \subset G$ which is what we wanted: to find a subcollection of $V_{\alpha}$ so that it is a subset of $G$. Then since $G$ and $k$ were arbitrary, $K$ must have a countable base.
To show that $K$ is dense can we say that since $\{V_{\alpha}\}$ is a finite open cover of $K$, then consider $G_{\alpha} = V_{\alpha} \cap K$. Then since $\{V_{\alpha}\}$ is a countable sequence, then so is $\{G_{\alpha}\}$. I think the union of $G_{\alpha}$ for every $\alpha$ is $K$. Then by definition is $K$ dense in $K$, and since it is countable, then $K$ is separable.
Otherwise, we can repeat how we solved problem $24$ just before this. Since $K$ is compact every infinite subset has a limit point in $K$.
The idea is slightly miscomunicated, at least. For each $n=1,2,\ldots$ we have open covers $$K=\bigcup\limits_{x\in K} B(x,n^{-1})$$
whence we may pick a finite subcover say $$B_1^{(n)}=B(x_1^{(n)},n^{-1}),\ldots,B_{k_n}^{(n)}=B(x_{k_n}^{(n)},n^{-1})$$
Note the number of open sets $k_n$ depends on $n$. Thus the collection of $B_j^{(i)}$ is a countable base and $$D=\{x^{(i)}_{j}:i=1,2,\ldots\; 1\leqslant j\leqslant k_i\}= \{x_1^{(1)},\ldots,x_{k_ 1}^{(1)},x_1^{(2)},\ldots,x_{k_2}^{(2)},\ldots\}$$ is a countable dense subset of $K$.
Can you prove this?