Show that every compact metrizable space has a countable basis.
My try:
Let $X$ be a compact space and metrizable. Now for each $n\in \Bbb N$; I can consider the open cover $\{B(x,\frac{1}{n}):x\in X\}$ of $X$ .As $X$ is compact we can find finite number of $x_i;1\leq i\le n$ corresponding to each $n$ .
$B_n=\{B(x_i,\frac{1}{n});1\leq i\le n\}$ .Now $\Bbb B=\{B_n:n\in \Bbb N\}$ is a countable collection .
It remains to show that $\Bbb B$ is a basis of $X$.
Let $U$ be an open set in $X$.Let $x\in U\implies \exists r>0$ such that $B(x,r)\in U$.Then we have for some $n;\frac{1}{n}<r\implies x\in B(x,\frac{1}{n})\subset B(x,r)\subset U$
The problem is I can't show the existence of a member of $\Bbb B$ say $B_n$ such that $x\in B_n\subset U$.
Can someone please help to complete the above proof ? I will be grateful if done so.