Separability and the Nagata-Smirnov Metrisation Theorem

It's comparable to the write-up in my note (note the similarity to 4.14 in Kelley):

Lemma 3: Let $X$ be regular and $T_0$ and let $d_i, i \in \mathbb{N}$ be a countable family of continuous pseudometrics on $X$ such that they are all bounded by $1$ and such that for every closed $A$ and every $x \notin A$, there is some $i$ such that $d_i(x,A)>0$, where $d_i(x,A)=\inf\{d_i(x,a): a \in A\}$ as usual. Then $X$ is metrisable and $$d(x,y)=\sum_i \frac{1}{2^i}d_i(x,y) $$ is a compatible metric.

So it's not an embedding argument but a direct construction of the metric from the countably many continuous pseudometrics. The countability of the family comes from the "$\sigma$" in $\sigma$-locally finite base, and the well-definedness of each pseudometric from the local finiteness, if you analyse the proof.

You can prove the lemma by the embedding argument, as Kelley does: let $X_n$ be $X$ in the topology induced by the pseudometric $d_n$, the continuity of the pseudometric implies that the identity mapping $i_n$ from $X$ to $X_n$ is continuous. A standard fact is that the weighted sum pseudometric induces the product topology on $\prod_n X_n$. The property of the family of pseudometrics then implies the family of $i_n$ separates points and closed sets and the standard embedding theorem then gives the stated result that the $d$ metric I defined works. So we don’t embed into a countable product of copies of reals but in a countable product of pseudometric spaces of the same size as $X$. Not separable necessarily.