If $X$ are a infinte countable set, define a metric in $X$ such that no point are isolated (i.e. for all $a\in X, \epsilon>0$ there exists $x\in B(a;\epsilon)\setminus\{a\})$.
I tried this but I don't know if it's correct: once $X$ is infinite countable then it can be put in 1-1 correspondence with $\mathbb{Q}$. Then we can write $X=(x_r)_{r\in\mathbb{Q}}$ and define distance in $X$ by $d(x_r,x_s) = |r-s|$. Once $\mathbb{Q}$ is dense in $\mathbb{R}$ no point will be isolated. Is this approach correct?
I just don't feel it, once the integers could fit in this with this euclidean norm, and still will not be a discrete metric space... Thanks in advance.
There is no "choice" but to go to $\mathbb{Q}$: every countable metric space without isolated points is homeomorphic to $\mathbb{Q}$. You can find proofs here.
So your idea works: take a bijection $f$ between the countable set $X$ and $\mathbb{Q}$ (which can be done as their both in bijective correspondence with, say $\mathbb{N}$) and define $d(x,x') = |f(x) - f(x')|$ where the right hand side has the absolute value as defined on the reals. Then the countable set has a metric that makes it isometric to $\mathbb{Q}$ in the usual distance (by definition basically), so these spaces are now homeomorphic so $X$ has no isolated points just like $\mathbb{Q}$ has none.