In Probability Theory by Klenke, the author proves the following lemma:
Let $f, g: \Omega\to E$ be a measurable function with respect to $\mathcal A-\mathcal B(E). $ Then the map $H: \Omega\to [0,\infty),~\omega \mapsto d(f(\omega), g(\omega)) $ us $\mathcal A-\mathcal B([0,\infty)$-measurable.
In the derivation, he first mentioned by triangle inequality
$$d(x,z)+d(z,y)\geq d(x, y) ~\forall x, y\in E; ~z\in F$$ where $F\subset E$ is countable and dense. Then he considered $\langle z_n\rangle_{n\in\mathbb N}$ be in $F$ such that $z_n\to x. $ Then he argued since $d$ is continuous, as $n\to \infty,$
$$d(x, z_n) +d(z_n, y) \to d(x, y) \tag 1\label 1$$ and putting things together
$$\inf_{z\in F}(d(x,z)+d(z,y)) = d(x, y) \tag 2\label 2.$$
What I am not able to see is how $\eqref 1$ happens: won't that be
$$\lim_{n\to \infty}(d(x, z_n) +d(z_n, y)) \geq d(x, y) $$ as dictated by the triangle inequality and subsequently
$$\inf_{z\in F}(d(x,z)+d(z,y)) \geq d(x, y)? $$
I know I am missing something. So any insight would be appreciated.
First, $z_n \to x$ means $d(x,z_n) \to 0$. Next, the sequential continuity of the function $d(-,y)$ yields that $$ z_n \to x \implies d(z_n,y) \to d(x,y). $$