There exists a counterexample, that is, a non-separable semi-metrizable space with ccc, which can be constructed as follows. Consider the countable power $\mathbb Z^\omega$ of the group of integers. For every $n\in\omega$ consider the "upper cone" $$V_n=\{x\in\mathbb Z^\omega:\forall i<n\;x(i)=0\mbox{ and }\forall i\ge n\;x(i)\ge0\}$$ and its symmetrization $X_n:=V_n\cup(-V_n)$.

On the group $\mathbb Z^\omega$ consider the topology $\tau$ consisting of sets $U\subset\mathbb Z^\omega$ such that for every $x\in U$ there exists $n\in\omega$ such that $x+X_n\subset U$.

The space $X=(\mathbb Z^\omega,\tau)$ (which is an Abelian quasi-topological group) has the required properties: it is semi-metrizable, non-separable and has ccc. The semi-metrizability of $X$ is witnessed by the function $g(n,x)=x+X_n$. The non-separability of $X$ follows from the fact that for every countable subset $A\subset \mathbb Z^\omega$ there is a function $f\in\mathbb Z^\omega$ such that $f\notin\bigcup_{a\in A}(a+X_0)$ and hence $A\cap(f+X_0)=\emptyset$.

To see that $X$ has ccc, take an uncountable family $\mathcal U$ of non-empty open sets. We can assume that each $U\in\mathcal U$ is of basic form $x_U+X_{n_U}$ for some $x_U\in\mathbb Z^\omega$ and $n_U\in\omega$. By the Pigeonhole Principle, there exists $n\in\omega$ such that the subfamily $\mathcal U_n=\{U\in\mathcal U:n_U=n\}$ is uncountable. Since the set $\mathbb Z^n$ is countable, we can apply the Pigeonhole Principle once more and find an element $x\in\mathbb Z^n$ such that the set $\mathcal U_{n,x}=\{U\in\mathcal U_n:x_U|n=x\}$ is uncountable. Finally, take any two distinct sets $U,V\in\mathcal U_{n,x}$ and observe that the intersection $U\cap V$ is not empty as it contains the functions $\max\{x_V,x_U\}$ and $\min\{x_V,x_U\}$.

shortfor CCC???? $\endgroup$ – bof Apr 21 '17 at 10:12