No. An example is the Arens-Fort Space.
Let $X = \omega \times \omega$. For each $A \subseteq X$ and $n \in \omega$ we let $A_n = \{ m \in \omega : (n,m) \in A \}$ denote the $n$th section of $A$.
Topologise $X$ by taking each point of $X \setminus \{ (0,0) \}$ to be isolated, and let $U \subseteq X$ be a neighbourhood of $(0,0)$ iff it contains $(0,0)$ and all but finitely many sections of $U$ are co-finite.
Clearly this space is T$_1$. If $F, E \subseteq X$ are disjoint closed sets, then one, say $E$, does not contain $(0,0)$. So $E$ is clopen, and $X \setminus E$ is an open set including $F$ which is disjoint from $E$. Thus $X$ is normal.
To show that $X$ is not metrizable, it suffices to show that it is not first-countable. For this we will show that there is no countable base at $(0,0)$. (Obviously, every other point has a countable base.) If $\{ U^{(i)} \}_{i \in \omega}$ is any family of open neighbourhoods of $(0,0)$, we inductively construct a sequence of pairs of natural numbers $\{ (n_i,m_i) \}_{i \in \omega}$ so that:
- $n_i > n_{i-1}$ is such that $U^{(i)}_{n_i}$ is non-empty (say $n_{-1} = 0$); and
- $m_i \in U^{(i)}_{n_i}$.
Then $V = X \setminus \{ ( n_i , m_i ) : i \in \omega \}$ is an open neighbourhood of $(0,0)$, and by construction $U^{(i)} \not\subseteq V$ for all $i$.