An interesting example is the subspace $X=A\cup B$ of the Niemitzki plane, where $A=\Bbb R\times \{0\}$ and $B=\{(a/n, 1/n):a\in\Bbb Z\land n\in\Bbb N\}.$ The subspace $A$ is closed in $X$ & discrete, with $|A|=2^{\aleph_0}.$ ... $B$ is a countable union of open singletons of $X$, so $B$ is also discrete. $B$ is dense in $X,$ so $X$ is not normal by the Jones Lemma. Any nbhd in $X$ of any $(r,0)\in A$ has an open-and-closed sub-nbhd $V=\{(r,0)\}\cup W$ where $W\subset B$ and $\overline W=V,$ so $V$ is homeomorphic to the real subspace $\{0\}\cup \{2^{-j}:j\in\Bbb N\}.$
In General Topology by R. Engelking, an example non-normal $Y=A\cup \Bbb N \subset \beta \Bbb N$ is given, where $A$ is a closed discrete subspace of $Y,$ with $A\subset \beta \Bbb N\setminus \Bbb N$ and $|A|=2^{\aleph_0}.$ I noticed that $Y$ is homeomorphic to $X$ of my 1st paragraph.
It seems the hypotheses of the Q are redundant. Wouldn't a locally metrizable space be $T_1$ and $T_{3\frac 1 2}$?
It would be interesting to see an example (if there is one) that does not involve the Jones Lemma.
BTW, in the Niemietzki plane, a basic open set $\{(r,0)\}\cup D,$ where $D$ is an open disc in $\Bbb R^+\times \Bbb R,$ of radius $s>0$ and centered at $(r,s),$ can also be seen to be metrizable by another strong metrization theorem: A $T_1,T_3$ space is metrizable iff it has a $\sigma$-locally finite base (basis). From this theorem, if $D$ is a metrizable open subspace of $E=D\cup \{p\}$ where $E$ is $T_1, T_3$ and $p$ has countable character in $E$, then $E$ is metrizable.
