Definition. Given a set $S$, a uniformity on $S$ is a filter $\mathcal{U}$ in $S\times S$ with the following properties:
(a) Every $A\in \mathcal{U}$ includes the diagonal $D:=\{\langle s,s\rangle:s\in S\}$.
(b) For each $A\in \mathcal{U}$, we have $A^{-1}:=\{\langle y,x\rangle:\langle x,y\rangle\in A\}\in \mathcal{U}$.
(c) For each $A\in \mathcal{U}$, there is a $B\in \mathcal{U}$ with $B\circ B\subset A$.
Theorem A uniformity $\mathcal{U}$ for a space $S$ is pseudometrizable (it is the pseudometric uniformity for some pseudometric $d$) if and only if $\mathcal{U}$ has a countable base.
Proof. "If": Let the uniformity $\mathcal{U}$ have a countable base {$U_n$}. For any $U$ in a uniformity $\mathcal{U}$, applying (c) twice, there is a $V\in \mathcal{U}$ with $(V\circ V)\circ(V\circ V)\subset U$. Recursively, let $V_0:=S\times S$. For each $n=1,2,...$,
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let $W_n$ be the intersection of $U_n$ with a set $V\in \mathcal{U}$ satisfying $(V\circ V)\circ(V\circ V)\subset V_{n-1}$.
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Let $V_n:=W_n\cap W_n^{-1}\in\mathcal{U}$. Then {$V_n$} is a base for $\mathcal{U}$, consisting of symmetric sets, with $V_n\circ V_n\circ V_n\subset V_{n-1}$ for each $n\geq 1$.
My question is: $V_n$ is intersection of $W_n$. $W_n$ is intersection of $U_n$. What's the usage of $V$ there?
The $U_n$ are a priori unrelated elements of the uniformity, and for the remainder of the proof (to construct the pseudometric) we need them to be sort of nicely decreasing "with room to spare", and symmetric too. So we construct $V_n \subseteq U_n$ that are still in $\mathcal{U}$ (so they remain a base for $\mathcal{U}$ ) but we want them to be symmetric and "decreasing with room to spare" in the sense that $V_n \circ V_n \circ V_n \subseteq V_{n-1}$ for all $n$. The construction goes in steps to show how the different axioms for a uniformity are used here:
$V_0$ is trivial, to get started (it's a member of any uniformity). For the recursion we assume we have $V_{n-1}\in \mathcal{U}$ (at stage $n$) and we want to find a $V_n$ as needed.
Then find $V' \in \mathcal{U}$ such that $(V' \circ V') \subseteq V_{n-1}$ by (c) applied to $V_{n-1}$ and then apply (c) to $V'$ and we have $V\in \mathcal{U}$ such that $V \circ V \subseteq V'$ and combining these we get
$$(V \circ V) \circ (V \circ V) \subseteq V_{n-1}\tag{1}$$
Then define $W_n = V \cap U_n \subseteq U_n \in \mathcal{U}$ (as entourages form a filter) which takes care of being a subset of $U_n$ but because $W_n \subseteq V$ we also have from (1):
$$(W_n \circ W_n) \circ (W_n \circ W_n) \subseteq V_{n-1}\tag{2}$$
which implies (we don't need four sets but only three, we can omit one because $\Delta_X \subseteq W_n$ by (a)):
$$W_n \circ W_n \circ W_n \subseteq V_{n-1}\tag{3}$$
But the $W_n$ is not yet symmetric, so we note that by (b) $W_n^{-1} \in \mathcal{U}$ too and $V_n := W_n \cap W_n^{-1}$ is symmetric (easy reasoning: $(x,y) \in V_n \to (x,y) \in W_n \land (x,y) \in W_n^{-1}$ so $(y,x) \in W^{-1}_n \land (y,x) \in W_n$ so $(y,x) \in V_n$) and $V_n \in \mathcal{U}$ as entourages are filters so closed under finite intersections. As $V_n \subseteq W_n$ we have
$$V_n \subseteq U_n \text{ and } V_n = V_n^{-1} \text{ and } V_n \circ V_n \circ V_n \subseteq V_{n-1}\tag{4}$$
and $V_n$ is our required set in the recursion. The proof makes all intermediate steps explicit (which is needed when you start uniformity theory), later you just say "we need some entourage and we can assume it's symmetric and small enough etc." but the above are the steps behind such reasoning. Hope it clarifies things.