Show that every compact metrizable space has a countable basis 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.

You need to redefine your $\Bbb B$. 
For each $n\in \Bbb N^+$ the corresponding open cover $B(x,1/n),x\in X$ admits a finite subcover 
$$B_n:=\{B(x_{ni},1/n),i=1,2,\cdots,k_n\}$$
due to compactness of $X$. 
Now you want to show $\Bbb B:=\cup B_n$ is a desired basis. 
Countableness has been shown in your attempt, it only suffices to show $\Bbb B$ is a basis indeed. That's to say, for all open subset $U\subset X$ and for all $x\in U$, there exists a member $B\in \Bbb B$ such that 
$$x\in B\subset U$$
Since there exists $r>0$ such that $B(x,r)\subset U$, we need only to find a $B$ that's completely included in $B(x,r)$. For $n$ so large that $1/n<r/4$, find some $m\in \{1,2,\cdots,k_n\}$ such that $x\in B(x_{nm},1/n)$, then it is readily apparent that this ball can serve as the desired $B$. (For all $y\in B(x_{nm},1/n)$, we have 
$$d(y,x)\le d(x,x_{nm})+d(y,x_{nm})<r/2<r$$
which says $B(x_{nm},1/n)\subset B(x,r)\subset U$.)
