Question on $T_1$-topological spaces with compatible uniformities having countable bases It is known that 

A $T_1$-topological space is metrizable $\iff$ it has a compatible uniformity with a countable base.

Let $(X,\tau')$ be a $T_1$-topological space and $\mathcal U'$ be a compatible uniformity on $X$ (i.e. $\tau_{\mathcal U'}=\tau')$ with countable base. Then by the above result $X$ is metrizable.
Now my question is 

Is it always possible to find a metric $d$ on $X$ such that $\mathcal U'$ is compatible with $d$ i.e. $$\mathcal U(d)=\mathcal U'?$$

 A: Once you have $\mathcal{U}(d) =\mathcal{U}'$, the topology induced by that metric
is the same as the topology induced by $\mathcal{U}'$, so $\tau'$.
This is because the intermediate step is superfluous in :
$d$ metric induces a uniformity, uniformity induces a topology.
is the same as metric induces that topology in one go, using the open balls.
This is clear when you look at the definitions of how we induce these topolgoies resp. uniformities:
Recall that a base for $\mathcal{U}(d)$ (as entourages) is all sets $\{(x,y) \in X^2: d(x,y) < \varepsilon\}$ and when we have a base for the entourage uniformity
$\mathcal{U}$ its induced topology on $X$ has as a base all sets $B[x]$ with $B$ in that base and $x \in X$, and for the standard $d$-uniformity base these $B[x]$ are just the open balls around $x$ with radius $\varepsilon$. 
Also, in the proof of the uniform metrisability theorem (Birkhoff-Kakutani?) we explicitly construct a compatible metric for the uniformity with a countable base, so in any scenario the answer is yes, you can.
