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.
 A: 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 $B(x_{nm},1/n)\subset B(x,r)\subset U$.)
A: First of all, Vim's comment is absolutely right: you might need more than $n$-many balls of radius ${1\over n}$ to cover your space. But of course this doesn't affect the argument, which only needs there to be finitely many. So let's let $x_{n, i}$ ($i\le k_n$) be the corresponding centers.
Hint: If $U$ is open and $x\in U$, then there is some $s$ such that $B(x, s)\subseteq U$. Now pick $n$ "large enough" and $x_{n, i}$ such that $x\in B(x_{n, i}, {1\over n})$; what can you say about $B(x_{n, i}, {1\over n})$ and $U$?

An interesting side note: this argument also shows that any compact metrizable space is separable (=has a countable dense subset), and it is easy to show that if a metrizable space is separable then it has a countable basis. In general topological spaces, this doesn't work: for instance the Stone-Cech Compactification of $\mathbb{N}$ (or, the usual topology on the set of ultrafilters on $\mathbb{N}$) has a countable dense subset but no countable basis. Metric spaces are special.
A: You have pretty much got it though there are some minor errors in your proof.
For each $n$ we consider the open cover $C_n=\{B(x, 1/n):\ x\in X\}$ of $X$. By compactness, there is a finite subset $X_n$ of $X$ such that $B_n:=\{B(x, 1/n):\ x\in X_n\}$ covers $X$.
(Here you made a mistake in your proof by implicitly assuming that $X_n$ can be chosen to have cardinality $n$).
Now we define $B=\bigcup_{n\in \mathbf N}B_n$.
(Here too there is a minor error in your proof. You define $B=\{B_n:n\in \mathbf N\}$. Note that the way I defined $B$ is different than yours.)
We want to show that $B$ forms a basis of $X$.
To show this, it suffices to show that every open set in $X$ can be written as a union of some members of $B$. So let $U$ be an open set in $X$ and $x\in U$ be chosen arbitrarily.
Then let $k>0$ be an integer such that $B(x, 1/k)$ is contained in $U$. Now by our construction, we know that $B_{2k}$ is an open cover of $X$. So some member of $B_{2k}$ contains $x$. There there is $x_0\in X_{2k}$ be such that $x\in B(x_0, 1/2k)$.
By triangle inequality, we have $B(x_0, 1/2k)\subseteq B(x, 1/k)\subseteq U$.
Thus we have found a member $B(x_0, 1/2k)$ of $B$ which contains $x$ and is contained in $U$, showing that $U$ can be written as a union of some of the members of $B$ and we are done.
A: I think you need a somewhat different proof.  For each $n$ consider $\{B(x,1/n):x\in X\}$.  Refine it to a finite cover using compactness.  The union of those finite covers for all $n$ is a countable set of balls.  Let the $x_i$ be the centers of those balls.  For every $x$ and every $n$, $x$ has to be in the finite cover corresponding to $n$, and thus there is an $x_i$ such that $d(x_i,x)<1/n$.  
Now, given $U$ and $x\in U$ choose $n$ such that $B(x,1/n)\subset U$, and choose $x_i$ such that $d(x_i,x)<1/(2n)$.  If $d(x_i,y)<1/(2n)$ then $d(x,y)\le d(x_i,x)+d(x_i,y)<2/(2n)=1/n$.  Thus $B(x_i,1/(2n))\subset B(x,1/n)\subset U$ as desired.
A: I am using slightly different notation, ie. $B^{\epsilon}(x) \equiv B(x; \epsilon)$.  For each $n$ we have that $B_n$ covers $X$.  Let $U$ be a neighborhood of $x$, hence $\exists \epsilon > 0$ such that $x \in B^{\epsilon}(x) \subset U$.  Let $n > 1/ 2\epsilon$ (or equivalently $2/n < \epsilon$). Since $B_n$ is a cover of $X$ we have that $x \in B^{1/n}(y)$ for some $y \in X$. All that remains to be shown is that this ball is contained within $B^{\epsilon}(x) \subset U$. Now, if we let $z \in B^{1/n}(y)$ then $d(x,z) \leq d(z,y) + d(x,z) = 1/n + 1/n < \epsilon$, meaning that $z \in B^{\epsilon}(x)$. So, we have that $x \in B^{1/n}(y) \subset B^{\epsilon}(x) \subset U$ and we are done.
