Understanding proof of: If a metric space contains a countable dense subset, then it satisfies the second axiom of countability Prop. If a metric space $(M, d)$ has a countable dense subset, then $M$ satisfies the second axiom of countability, i.e. $M$ has a countable base.
The proof is as follows (the part in blockquote is the part I do not understand and wasn't able to do myself):
Suppose $M$ has a countable dense subset i.e. $A = \{ x_{1}, x_{2}, \ldots, x_{n}, \ldots \}$. This means that closure $cl(A) = M$. We want to show that there is a system $\mathscr{B}$ of subsets of $M$ such that for each open set $U$ and each $x \in U$ there is a $B \in \mathscr{B}$ such that $x \in B$ and $B \subseteq U$.

Now the idea is to take $\mathscr{B}$ as the set of open balls $B(x_{m}, \frac{1}{n})$. This is justified by the author like this: Given an open set $U \subseteq M$ and any $x \in U$ there is an open ball $B(x_{m}, \frac{1}{n})$ such that $x \in B(x_{m}, \frac{1}{n}) \subseteq U$ for suitable positive integers $m$ and $n$.

My question is why is there for each open set $U$ and for any $x \in U$ an open ball $B(x_{m}, \frac{1}{n})$ such that $x \in B(x_{m}, \frac{1}{n}) \subseteq U$ for suitable positive integers $m$ and $n$? And what does suitable mean here?
 A: $\{x_1,...,x_n,...\}$ is dense is equivalent to the fact that for every $x$ there exists $x_{n_q}$ a subsequence of $(x_n)$, such that $limx_{n_q}=x$. Since $U$ is open, there exists $B(x,r)\subset U$, take $n_q$ such that $d(x,x_{n_q})\leq r/4$ and $m$ such that $1/m<r/4$, for every $y\in B(x_{n_q},1/m), d(x,y)\leq d(x,x_{n_q})+d(x_{n_q},y)\leq r/4+r/4<r$ implies that $y\in B(x,r)$ and $B(x_{n_q},1/m)\subset U$.
A: Since A is dense and U open, there is a x in A with x in U.
As U is open, exists n in N with B(x,1/n) subset U.  
This allows you show U is a countable union of such balls.
A base for the space is accordingly
{ B(x,1/n) : x in A, n non negative integer }.
A: First note that given $x$ and $U$ open, there is an $r>0$ such that $B(x,r) \subseteq U$, by the definition of a topology induced by the metric. Take $n$ large enough so that $\frac{1}{n} < \frac{r}{2}$. Note that then $B(x,\frac{1}{n})$ must intersect the dense set $A$ (as $x$ is a limit point of $D$) say $x_m \in B(x, \frac{1}{n})\cap A$. So $d(x_m,x) < \frac{1}{n}$ which also says $x \in B(x_m, \frac{1}{n})$
I claim that $B(x_m, \frac{1}{n}) \subseteq U$: If $y \in B(x_m, \frac{1}{n})$ then $$d(x,y) \le d(x,x_m)+d(x_m,y) < \frac{1}{n} + \frac{1}{n}< \frac{r}{2}+\frac{r}{2}=r$$
so $y \in B(x,r)$ and so $y \in U$, proving the inclusion claim.
linguistic remark: "for suitable integers $m,n$" is just a way of saying that "$\exists m,n \in \Bbb N$ such that". The existence of the $m,n$ follow from the above proof: the $n$ I can choose because  $\frac{1}{n}$ can be made as small as I like by picking $n$ large enough (you should know that at least). The $m$ exists because it's the index of the element of the countable set $A$ that intersects $B(x, \frac{1}{n})$.
