= Solution
If $\mathcal B$ is a <countable> <topological basis> of $X$, then $\{B\cap Y:B\in\mathcal B\}$ is a <countable> <topological basis> of the subspace $Y$. Indeed every subspace-<open neighborhood> is $Y\cap U$ for an open $U$ of $X$, and a suitable basic set lies between its <point> and $U$. This proves the first assertion.
\b[<Urysohn's metrisation theorem>:] every <second-countable space> that is also a <T3 space> is a <metrizable space>. Here <regularity of a topological space> allows a <neighborhood> of a <point> to be shrunk to an <open neighborhood> whose <closure> is still inside the original one.
We first prove the separation facts needed in a complete proof. <Second countability> implies the <Lindelöf> property: for each basic set contained in some member of an <open cover> choose one such member; these countably many members cover. In particular closed subspaces are <Lindelöf>. For disjoint <closed sets> $F,G$, <regularity of a topological space> and this property give open covers $F\subseteq\bigcup_nU_n$, $G\subseteq\bigcup_nV_n$ with $\overline U_n\cap G=\varnothing$ and $\overline V_n\cap F=\varnothing$. Define
$$
U=\bigcup_n\left(U_n\setminus\bigcup_{j\leq n}\overline V_j\right),\qquad V=\bigcup_n\left(V_n\setminus\bigcup_{j\leq n}\overline U_j\right).
$$
These sets are open and contain $F,G$ respectively. They are disjoint: if a <point> belonged to a term indexed by $n$ in the first union and by $m$ in the second, the term with the larger index would have removed the <closure> of the other covering set. Thus the space is a <normal space>.
For completeness, <Urysohn's lemma> supplies <continuous> separating functions in a <normal topological space>. Its construction is as follows. Given disjoint closed $F,G$, choose <open sets> $O_r$ for dyadic $r\in[0,1]$, with $F\subseteq O_0$, $O_1=X\setminus G$, and $\overline O_r\subseteq O_s$ whenever $r<s$. Start by shrinking a <neighborhood> of $F$ inside $X\setminus G$, and insert the successive dyadic sets by the <closed-neighborhood shrinking in a normal space>. Define
$$
f(x)=\inf\bigl(\{r:x\in O_r\}\cup\{1\}\bigr).
$$
Then $f=0$ on $F$, $f=1$ on $G$, and $0\leq f\leq1$. The identities $\{f<a\}=\bigcup_{r<a}O_r$ and $\{f>a\}=\bigcup_{a<r<1}(X\setminus\overline O_r)$ prove <continuity> for $0<a<1$, with the endpoint cases immediate.
Let $\{B_n\}$ be a <countable> <topological basis>. For each pair $(n,m)$ with $\overline B_n\subseteq B_m$, choose a <continuous function> $f_{nm}:X\to[0,1]$ equal to one on $\overline B_n$ and zero outside $B_m$. These form a <countable> family. Given $x\in O$ open, choose $B_m$ with $x\in B_m\subseteq O$, shrink a <neighborhood> of $x$ inside $B_m$, and then choose $B_n$ inside that smaller <neighborhood>. Thus
$$
x\in\{f_{nm}>1/2\}\subseteq B_m\subseteq O.
$$
Consequently these functions generate the original topology. They also separate distinct <points>: use the $T_1$ <open set> $X\setminus\{y\}$ when $x\ne y$. Enumerate them as $f_j$, padding by zero functions if necessary. The map $x\mapsto(f_j(x))_j$ is a <topological embedding> into the <countable> cube. Pulling back its <metric>, or directly using
$$
\boxed{d(x,y)=\sum_{j=1}^{\infty}2^{-j}|f_j(x)-f_j(y)|,}
$$
gives a compatible <metric>. The fact that the functions distinguish distinct points proves positivity for distinct <points>, the <triangle inequality> is termwise, and the uniformly bounded tails show that this <metric> induces exactly the topology generated by the coordinates. Empty and singleton spaces are trivially <metrizable spaces>. This proves the theorem.
For the application, a <separable> <metric space> has a <countable> <topological basis> consisting of balls with rational positive radii about a <countable> <dense> set, and it is a <T3 space>. The given product theorems make a <countable> product of these spaces <second countable> and a <T3 space>. Urysohn's theorem makes it a <metrizable space>. Finally every <second countable> space is <separable>: choose a <point> from each nonempty basic <open set>; the resulting <countable> set meets every nonempty <open set>. Hence the product is \b[<separable> and a <metrizable space>].
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