A sufficient condition for a collection of Borel probability measures to be uniformly tight

Question

I'm trying to prove below result. Could you verify if my attempt is fine?

Let $(X, d)$ be a complete metric space and $\mathcal{P}(X)$ the space of Borel probability measures on $X$. Let $\Gamma \subset \mathcal{P}(X)$. If $$ \forall \varepsilon, \delta>0, \exists a_{1}, \ldots, a_{n} \in X, \forall \mu \in \Gamma: \mu\left(\bigcup_{i=1}^{n} B\left(a_{i}, \delta\right)\right) \geq 1-\varepsilon, $$ then $\Gamma$ is uniformly tight.

I post my proof separately as below answer. This allows me to subsequently remove this question from unanswered list.

Answer 1

Fix $\varepsilon>0$. For each $m \ge 1$, there exists a finite set $A_m := \{a_{m, 1}, \ldots, a_{m,k_m}\} \subseteq X$ such that $$ \mu \left( \bigcup_{i=1}^{k_m} B \left(a_{m,i}, 1/m\right) \right) \geq 1- \varepsilon2^{-m} \quad \forall \mu \in \Gamma. $$

Let $B_m := \bigcup_{i=1}^m A_m$. We re-label $B_m$ such that $B_m = \{a_1, \ldots, a_{k_m}\}$. Let $$ K := \bigcap_{m=1}^\infty \bigcup_{i=1}^{k_m} \overline B \left(a_{i}, 1/m\right). $$

Clearly, $K$ is closed and thus complete. Moreover, $K$ is totally bounded, so it's compact. We have

$$ \mu(K^c) \le \varepsilon \sum_{m=1}^\infty 2^{-m} = \varepsilon \quad \forall \mu \in \Gamma. $$

This completes the proof.