Let $N$ be a countable indexing set, and $((X_k,\tau_k))_{k \in N}$ topological spaces. Define $X = \prod_{k \in N} X_k = X^N$ and let $\tau$ be the product topology on $X$ induced by $\tau_k$s.
Theorem: If all $(X_k,\tau_k)$s are first/second countable, then so is $(X,\tau)$.
Proof: For each $k \in N$ and $x \in X_k$, define $\mathcal{E}_k(x):\mathbb{W} \to \mathcal{P}(X)$ be a countable neighbourhood base(local basis) at $x$. Let $f \in X$. Define $$ \mathcal{F} = \left\{ \prod_{k \in N} \mathcal{E}_k(f(k))(\phi(k)) \mid \phi:N \to \mathbb{W} \right\} $$ ($\mathcal{E}_k(f(k))(\phi(k))$ means the mapping $\mathcal{E}_k \circ f(k)$ "evaluated at" $\phi(k)$.) Then sets in $\mathcal{F}$ are the intersection of cylinders $$ \bigcap_{k \in N} \pi_k^{-1} \circ \mathcal{E}_k(f(k))(\phi(k)) $$ Since the $\pi_k^{-1}$ are made continuous by the product topology, $\mathcal{F} \subseteq \tau$, but how do I show the two following properties?
$\forall S \in \mathcal{F}, f \in S$
$U \in \tau \land f \in U \rightarrow \exists S \in \mathcal{F} \mid S \subseteq U$
I think the argument for second countability is similar and follows directly from this one.