A topology that is finer than a metrizable topology is also metrizable? If $\tau_{1}$ and $\tau_{2}$ are two topologies on a set $\Omega$ such that $\tau_{1}$ is weaker than $\tau_{2}$ (i.e. $\tau_{1}\subset\tau_{2}$) and $\tau_{1}$ is metrizable, is it then true that $\tau_{2}$ is also metrizable? 
My guess is that it is not true, since we do not know what the sets in $\tau_{2}$ look like, let alone whether they can contain 'open metric balls' or not. But maybe it is possible to adapt the metric so that the sets in $\tau_{2}$ can contain 'open metric balls'.
Also, it may be the case that the converse is easier to work with: If $\tau_{2}$ is not metrizable, then $\tau_{1}$ is not metrizable.
I find it very hard to think of any (counter)examples. Any help or hints would be greatly appreciated!
 A: No, take $X=\Bbb R$ and $\tau_1$ the usual topology, clearly metrisable, and $\tau_2$ the Sorgenfrey (aka "lower limit") topology (generated by the sets of the form $[a,b)$). Then $\tau_1 \subsetneq \tau_2$ but $\tau_2$ is not metrisable, for several reasons, the most accessible of which are (and the reason it's covered in some many topology text books and papers): its square is not normal, or it's separable but it doesn't have a countable base. See Wikipedia for more info.
Another example is $\mathbb{R}^\omega$ in the (metrisable) product topology and the finer box topology (which is not even first countable).
Also Munkres' $\Bbb R_K$ space which is $\Bbb R$ in the usual topology but with one extra closed set $K=\{\frac{1}{n}: n \in \Bbb N^+\}$, is not even regular (and all metrisable spaces are normal and perfectly normal) is another elementary example (see 2nd edition, p.197/198.).
A: Additional counterexamples can be found in functional analysis. Let $X$ be a (real or complex) Banach space, and let $\mathcal T_{\text{norm}}$ denote the norm topology on $X$. If $X$ is infinite-dimensional, then there exist linear topologies $\mathcal T$ on $X$ that are complete and strictly finer than $\mathcal T_{\text{norm}}$, for example:


*

*One may take $\mathcal T$ to be the finest locally convex topology on $X$ — see [Sch99, p. 56, example before 6.3].

*For additional examples, see this answer on MathOverflow.


Such a topology $\mathcal T$ can never be metrizable, for otherwise it would follow from the open mapping theorem (cf. [Rud91, Corollary 2.12(d)]) that $\mathcal T = \mathcal T_{\text{norm}}$.

Added later: you don't even need the open mapping theorem for this — it can be shown by elementary means that the finest locally convex topology on an infinite-dimensional vector space is never metrizable (see the answers to this question).
Every infinite-dimensional vector space admits a norm (see here), and can therefore be equipped with two locally convex topologies $\mathcal T_{\text{norm}} \subset \mathcal T_{\text{finest lc}}$, where $\mathcal T_{\text{norm}}$ is metrizable but $\mathcal T_{\text{finest lc}}$ is not.

References.
[Rud91]: Walter Rudin, Functional Analysis, Second Edition (1991), McGraw–Hill.
[Sch99]: H.H. Schaefer, M.P. Wolff (translator), Topological Vector Spaces, Second Edition (1999), Graduate Texts in Mathematics 3, Springer.
A: A countable example is furnished by the subspace $X=\{0\}\cup\{2^{-n}:n\in\Bbb N\}$ of $\Bbb R$ with topology $\tau_0$ that it inherits from the usual topology on $\Bbb R$. Let $\mathscr{U}$ be a free ultrafilter on $\Bbb N$, and let $\tau=\{U\subseteq X:0\notin U\text{ or }\{n\in\Bbb N:2^{-n}\in U\}\in\mathscr{U}\}$; it is easily verified that $\tau$ is a topology on $X$ that is finer than $\tau_0$, but $\langle X,\tau\rangle$ is not even first countable.
