I'm in high school and self-studying analysis. I completed this proof for a problem in Rudin, but wanted some verification. Does this look correct?
Proof that every compact metric space $K$ has a countable base and is therefore separable:
Consider $p \in K$ with an arbitrary $\epsilon > 0.$ By the Archemedian Principle, there exists a natural number $n$ such that $\frac{1}{n} < \epsilon.$
Consider the open cover $$K \subset \bigcup_{i \in K} N_{\frac{1}{n}}(i).$$ Since $K$ is compact, there exists a finite subcover $$K \subset \bigcup_{i \in X_n} N_{\frac{1}{n}}(i),$$ where $X_n = \{x_{1_n}, x_{2_n}, x_{3_n}, … x_{j_n}\} \subset K$ for some $j_n \in \mathbf{N}.$
Since $p \in K$, $p \in N_{\frac{1}{n}}(x_{i_n})$ for some $x_{i_n} \in X_n$.
This means that $d(p, x_{i_n}) \leq \frac{1}{n}$, which also implies that $x_{i_n} \in N_\frac{1}{n}(p) \subset N_{\epsilon}(p),$ so $x_{i_n} \in N_{\epsilon}(p).$
Since $x_{i_n}$ is a member of $X_n$, $N_{\epsilon}(p)$ thus intersects the finite subset $X_n$.
As the choices for $p \in K$ and $\epsilon$ were arbitrary, any neighborhood around every point in $K$ must intersect a countable subset $X$.
By definition of closure of a set, $K \subset$ cl$(X)$. This means that $X$ is dense in $K$ and thus forms a countable base.
Therefore $K$ is separable.
Best Answer
The idea is correct, but not well argumented, I think. It's mostly a problem of notation, however, plus a weakness I'll underline later on.
I'd follow the hint, that is, proving first the space has a countable basis.
For all integers $n>0$, the open cover $\{N_{1/n}(p):p\in K\}$ has a finite subcover; let $X_n=\{x_{n,1}, x_{n,2}, \dots, x_{n,m(n)}\}$ be such that $$ K=\bigcup_{i=1}^{m(n)}N_{1/n}(x_{n,i}) $$ I claim that the set $$ \mathcal{B}=\bigcup_{n>0}\bigl\{N_{1/n}(x_{n,i}):1\le i\le m(n)\bigr\} $$ is a countable basis for $K$. Countability is obvious. Let $p\in K$ and $\varepsilon>0$; we want to prove that there exist $n>0$ and $i$ with $1\le i\le m(n)$ such that $N_{1/n}(x_{n,i})\subseteq N_\varepsilon(p)$.
Take $n$ such that $1/n<\varepsilon/2$. Then $p\in N_{1/n}(x_{n,i})$, for some $1\le i\le m(n)$. If $q\in N_{1/n}(x_{n,i})$, then $$ d(p,q)\le d(p,x_{n,i})+d(x_{n,i},q)<\frac{1}{n}+\frac{1}{n}<\varepsilon $$ so $N_{1/n}(x_{n,i})\subseteq N_\varepsilon(p)$ (this is a point where your proof is weak).
Now every metric space having a countable basis is separable. It suffices to take a point in each (nonempty) member of the basis and this is a dense subset, because each open set is the union of members of the basis, so it intersects this countable subset.