If $X=\beta\mathbb{N}$ and $M$ is countable then $X_M$ is a metric space. (Elementary submodels) I imagine this question is very simple but there must be something that I'm missing.
If $M$ is an elementary submodel of the universe, and I have a topological space $(X,\tau)\in M$. I can define a new topological space $X_M = (X\cap M, \tau\cap M)$. (The set $ \tau\cap M$ is not necesarrily a topology but it is a basis for one.)
It supposedly clear that if $M$ is countable and $X$ is $\beta\mathbb{N}$ then $(X\cap M, \tau\cap M)$ is a countable metric space. 
I don't know how to verify the claim. I know it is countable and second countable, but I do not know if it is compact or something else like this that will give me a metric.
Thank you for any help. 
 A: I believe the following works (it's a bit late where I am, but here goes):
First of all, let's clear up what the space is. As Asaf points out, the base is a bit messier than you've described: the space $(X_M, \tau_M)$ is $(X\cap M, \{U\cap M: U\in\tau\cap M\})$. Right off the bat, this tells us that if $M$ is countable then $(X_M,\tau_M)$ is a countable second-countable space, regardless of what $(X,\tau)$ is.
Now we want to argue that if $X=\beta\mathbb{N}$ and $\tau$ is as usual, the space $\mathcal{X}_M=(X_M,\tau_M)$ will be $T_3$; by Urysohn's metrization theorem (as observed by Henno Brandsma in a comment), the result will follow. So we need to show that $\mathcal{X}_M$ is Hausdorff and regular.


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*$\mathcal{X}_M$ is Hausdorff: given distinct ultrafilters $p, q\in X\cap M$, there must be a set distinguishing them in $M$ - that is, some $a\subseteq\omega$ with $a\in p$ but $\omega\setminus a\in q$. But then the open sets $[a]$ and $[\omega\setminus a]$ (where "$[x]$" denotes the set of ultrafilters containing $x$) forms a pair of open sets in $\mathcal{X}_M$ separating $p$ and $q$.

*$\mathcal{X}_M$ is regular: we want to show that if $p\in X_M$ and $C\subseteq X_M$ is closed, then there are neighborhoods separating $p$ and $C$. Since the complement of $C$ is open, we can find some basic open set containing $p$ which is disjoint from $C$; considering the topology, what this means is that we can find some $a\subseteq\omega$ with $a\in M$ and $x\in p$ (i.e. $p\in [a]$) but with $a\not\in q$ for any $q\in C$. But then $[\omega\setminus a]$ is a basic open neighborhood containing $C$ which is disjoint from $[a]$ (which contains $p$).
