First, find a fundamental system of entourages $U_0 \supseteq U_1 \supseteq \cdots$ such that

- $U_0 = X \times X$
- $(x_0,x_1) \in U_i$ iff $(x_1,x_0) \in U_i$.
- If $(x_0,x_1),(x_1,x_2),(x_2,x_3) \in U_{i+1}$ then $(x_0,x_3) \in U_i$.

Define $e(x_0,x_1) = \inf\lbrace 2^{-i} : (x_0,x_1) \in U_i\rbrace$. This is a well-defined nonnegative number since $(x_0,x_1) \in U_0$. The function $e$ is symmetric by property 2 and, of course, $e(x,x) = 0$. However, it does not necessarily satisfy the triangle inequality just yet.

By induction on $n$, we show that $$\sum_{i=0}^{n-1} e(z_i,z_{i+1}) \geq \frac{e(z_0,z_n)}{2}$$ for all sequences $z_0,\dots,z_n$. This is clear when $n = 1$. Suppose the result is true for all $n \lt m$. Given $z_0,\dots,z_m$, set $s = \sum_{i=0}^{m-1} e(z_i,z_{i+1})$.
We may assume $0 \lt s \lt \frac12$, otherwise the result is trivial. Let $n \lt m$ be such that $$\sum_{i=0}^{n-1} e(z_i,z_{i+1}) \leq \frac{s}{2} \lt \sum_{i=0}^n e(z_i,z_{i+1}).$$ By the induction hypothesis, $e(z_0,z_n) \leq s$ and also $e(z_{n+1},z_m) \leq s$ since $$\sum_{i=n+1}^{m-1} e(z_i,z_{i+1}) = s - \sum_{i=0}^{n} e(z_i,z_{i+1}) \leq \frac{s}{2}.$$ Clearly, $e(z_n,z_{n+1}) \leq s$ too. Let $k$ be such that $2^{-k-1} \leq s \lt 2^{-k}$. Then $(z_0,z_n),(z_n,z_{n+1}),(z_{n+1},z_m) \in U_{k+1}$ which implies that $(z_0,z_m) \in U_k$ and hence $e(z_0,z_m) \leq 2^{-k} \leq 2s$, as required.

If we define $$d(x,y) = \inf \sum_{i=0}^{n-1} e(z_i,z_{i+1})$$ where the infimum is taken over all finite sequences $z_0,\dots,z_n$ with $z_0 = x$ and $z_n = y$, then $d$ is a pseudometric and since we always have $$\frac{e(x,y)}{2} \leq d(x,y) \leq e(x,y),$$ this pseudometric generates the correct topology.