Let $\mathcal{T}$ be the topology on $\mathbb{R}$ generated by the standard topology and the excluded point topology. Explicitly, $\mathcal{T}=\{E\subset \mathbb{R}|(-\epsilon,\epsilon) \subset E$ for some $\epsilon>0 \} \cup \{E \subset \mathbb{R}| 0 \not \in E\}$ (where I have chosen $0$ to be the excluded point). I am curious to know whether $\mathcal{T}$ is metrizable.
If my calculations are correct, $\mathcal{T}$ is normal ($T_4$), first countable but not second countable, and not separable. Also $\mathcal{T}$ restricted to any subset not containing $0$ is the discrete topology, which is metrizable. $\mathcal{T}$ restricted to an arbitrary subset containing $0$ is harder to analyze, but when restricted to an open interval containing $0$ it is homeomorphic to $(\mathbb{R},\mathcal{T})$ itself, and when restricted to $\{0\} \cup \{\frac{1}{n}\}$ it is the same as that induced from the standard topology of $\mathbb{R}$, which is obviously metrizable. Hence it passes all the necessary conditions for metrizability that I am familiar with (and can get to apply).
On the other hand, $d(0,x)=\frac{|x|}{1+|x|}$ for $x\neq 0$, $d(x,y)=1$ for $x,y$ both $\neq0$ would induce $\mathcal{T}$ if $d$ were a metric, but it is not: $\frac{2}{3}=d(0,-\frac{1}{2})+d(\frac{1}{2},0)<d(\frac{1}{2},-\frac{1}{2})=1$. It seems possible that a clever way to modify $d$ may create a metric that works, but I cannot find one myself.
Any help is appreciated!
Hint: Keep your $d(x,0)$, and for distinct $x,y\neq 0$, set $d(x,y)=d(x,0)+d(0,y)$.