Every linear topological space embeds into the Tychonoff product of linear metric spaces I need a reference to the following (known?)

Fact. Every topological vector space $X$ over the field of real numbers is topologically isomorphic to a linear subspace of the Tychonoff product of linear metric spaces.

This seems to be a basic fact in the theory of topological vector spaces, but I did not find it in the classical textbooks: Bourbaki, Schaefer, Rolewicz, Robertson&Robertson. If you know a suitable reference, please help!
 A: 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.
A: In Jarchow's Locally Convex Spaces Theorems 2.7.3 and 2.9.2 on pages 39 and 43 together say that a Hausdorff topological vector space is linearly homeomorphic to a dense subspace of a projective limit of a projective system of metrizable topological vector spaces. Noting that in section 2.6 (p. 37) Jarchow defines the projective limit as a certain subspace of the product, this contains the required assertion provided that the space is Hausdorff.
A: This is wildly off topic but I couldn‘t resist giving the complete picture.
Every tvs (lcs, uniform space) is naturally isomorphic to a subspace of a product of $F$ spaces (complete metrisable ones) (Banach spaces, complete metric spaces).  Complete ones are isomorphic to closed subspaces in all three cases (an if and only if condition, of course). A tvs (lcs, uniform space) is emeddable as a closed subspace of a product of metrisable tvs‘s (normed spaces, metrisable spaces) if and only it satisfies a weaker completeness condition which I will make explicit in the case of uniform spaces——-a net $(x_\alpha)$ which satisfies the following condition is convergent: for each continuous semi-metric $d$ there is a $\gamma$ so that $d(x_\alpha,x_\beta)=0$ whenever $\alpha,\beta \geq \gamma$.
