Lemma 1: If $\{U_1, U_2,\ldots\}$ is a countable cover of $X$ and $\Gamma$ relatively compact, then
$$
\forall \varepsilon>0, \exists k \ge 1, \forall \mu \in \Gamma: \mu \left ( \bigg ( \bigcup_{i=1}^k U_i \bigg )^c \right ) < \varepsilon.
$$
Proof: Assume the contrary. Let $F_k := ( \bigcup_{i=1}^kU_i )^c$ .Then
$$
\exists \varepsilon>0, \forall k \ge 1, \exists\mu_k \in \Gamma: \mu_k \left (F_k \right ) > \varepsilon.
$$
There is a subsequence $(\mu_{\varphi (m)})$ and $\bar \mu \in \overline \Gamma$ such that $\mu_{\varphi (m)} \to \bar \mu$ in $d_P$. This implies $\mu_{\varphi (m)} \to \bar \mu$ weakly. Notice that $F_k$ is closed, so
$$
\varepsilon \le \lim\sup_m \mu_{\varphi (m)} \left ( F_k \right ) \le \bar \mu \left ( F_k \right ) \quad \forall k \ge 1.
$$
Notice that $F_k \searrow \emptyset$, so $\varepsilon \le 0$ which is a contradiction.
Fix $\varepsilon >0$. Let $D := \{a_1, a_2, \ldots\}$ be a countable dense subset of $X$. By Lemma 1, for each $m\ge 1$, there is $k_m$ such that
$$
\mu \left( \left [ \bigcup_{i=1}^{k_{m}} B\left(a_{i}, 1 / m \right) \right ]^c \right) < \varepsilon 2^{-m} \quad \forall \mu \in \Gamma.
$$
Take
$$
K := \bigcap_{m=1}^\infty \bigcup_{i=1}^{k_{m}} \overline B\left(a_{i}, 1 / m \right).
$$
Then $K$ is closed and totally bounded. Hence $K$ is complete and totally bounded. So $K$ is compact. Moreover,
$$
\mu(K^c) \le \varepsilon \sum_{m=1}^\infty 2^{-m} = \varepsilon \quad \forall \mu \in \Gamma.
$$
It follows that $\Gamma$ is tight.
