I don't have it at hand but I would look at Köthe's book *Topological Linear Spaces*.

I believe that a proof can be given in the following way: For each $0$-neighbourhood $U$ choose a decreasing sequence of $0$-neighbourhoods $U_n$ with $U_0\subseteq U$, $U_n+U_n\subseteq U_{n-1}$ and $\lambda U_n \subseteq U_{n}$ for all $|\lambda|\le 1$. Then the topology $\tau_U$ on $X$ having $(U_n)_{n\in\mathbb N}$ as a basis of the $0$-neighbourhood filter is semi-metrizable by the Birkhoff-Kakutani theorem (at least Wikipedia tells it like that). Then $X$ is isomorphic to the diagonal in the product of all ($X,\tau_U)$. If $X$ is Hausdorff you can pass to the associated Hausdorff spaces $X_U$ of the $(X,\tau_U)$ by factoring out $\overline{\{0\}}^{\tau_U}$. If $q_U:X\to X_U$ is the quptient map you still get an embedding $X\to \prod_U X_U$, $x\mapsto(q_U(x))_{U}$. You can also take the completions of $X_U$ to make $X$ isomorphic to a subspace of the product of complete linear metric spaces.

EDIT. The result is contained in $3 (4) of the *Springer Lecture Notes in Mathematics* 639, *Topological Vector Spaces* by Adasch, Ernst, and Keim.