Lemma 1: Let $f:X \to \mathbb R \cup \{+\infty\}$ be lower semi-continuous and bounded from below. Then there is an increasing sequence $(f_n)$ of $n$-Lipschitz continuous functions such that $f_n \nearrow f$ pointwise.
Lemma 2: Let $X,Y$ be nonempty and $f:X\times Y \to \mathbb{R}$. Then$$\sup_{y\in Y} \inf_{x\in X}f(x,y) \leq \inf_{x\in X} \sup_{y\in Y}f(x,y).$$
Lemma 3: Let $X,Y$ be nonempty and $f:X\times Y \to \mathbb{R}$. Then$$\sup_{y\in Y} \sup_{x\in X}f(x,y) = \sup_{(x,y) \in X \times Y} f(x,y) = \sup_{x\in X} \sup_{y\in Y}f(x,y).$$
Let's prove the first direction. Let $\mu, \mu_1, \mu_2, \ldots \in \mathcal P(X)$ such that $\mu_n \to \mu$ weakly. Fix a lower semi-continuous function $f:X \to \mathbb R \cup \{+\infty\}$ that is bounded from below. WLOG, we assume $f$ is bounded by $0$. By Lemma 1, there is an increasing sequence $(f_m)$ of $m$-Lipschitz continuous functions such that $0\le f_m \nearrow f$ pointwise. Then
\begin{align}
\liminf_n \int f \mathrm d \mu_n &= \liminf_n \int \sup_m f_m \mathrm d \mu_n \\
&= \liminf_n \sup_m \int f_m \mathrm d \mu_n \quad \text{by monotone convergence}\\
&= \sup_n \inf_{k\ge n} \sup_m \int f_m \mathrm d \mu_k\\
&\ge \sup_n \sup_m \inf_{k\ge n} \int f_m \mathrm d \mu_k \quad \text{by Lemma 2} \\
&= \sup_m \sup_n \inf_{k\ge n} \int f_m \mathrm d \mu_k \quad \text{by Lemma 3} \\
&= \sup_m \lim_n \int f_m \mathrm d \mu_n\\
&= \sup_m \int f_m \mathrm d \mu \quad \text{by weak convergence} \\
&= \int \sup_m f_m \mathrm d \mu \quad \text{by monotone convergence} \\
&= \int f \mathrm d \mu.
\end{align}
Let's prove the reverse direction. Fix a continuous bounded $f:X \to \mathbb R$. Clearly, $f$ and $-f$ are l.s.c. and bounded from below. So
$$
\liminf_n \int f \mathrm d \mu_n \ge \int f \mathrm d \mu \quad \text{and} \quad \liminf_n \int -f \mathrm d \mu_n \ge \int -f \mathrm d \mu.
$$
Then
$$
\liminf_n \int f \mathrm d \mu_n \ge \int f \mathrm d \mu \quad \text{and} \quad -\limsup_n \int f \mathrm d \mu_n \ge -\int f \mathrm d \mu.
$$
This completes the proof.
Update: I have found another approach below.
Let's prove the first direction. Let $\mu, \mu_1, \mu_2, \ldots \in \mathcal P(X)$ such that $\mu_n \to \mu$ weakly. Fix a lower semi-continuous function $f:X \to \mathbb R \cup \{+\infty\}$ that is bounded from below. WLOG, we assume $f$ is bounded by $0$.
By Lemma 1, there is an increasing sequence $(f_m)$ of $m$-Lipschitz continuous functions such that $0\le f_m \nearrow f$ pointwise. Let $g_m := \min \{f_m, m\}$. Then $(g_m)$ is a sequence of continuous bounded functions such that $0\le g_m \nearrow f$ pointwise. We have
$$
\int f \mathrm d \mu_n \ge \int g_m \mathrm d \mu_n \quad \forall m \in \mathbb N.
$$
Then
$$
\liminf_n \int f \mathrm d \mu_n \ge \liminf_n \int g_m \mathrm d \mu_n = \int g_m \mathrm d \mu \quad \forall m \in \mathbb N.
$$
Then
$$
\begin{align}
\liminf_n \int f \mathrm d \mu_n &\ge \liminf_m \int g_m \mathrm d \mu \\
&\ge \int \liminf_m g_m \mathrm d \mu \quad \text{by Fatou Lemma} \\
&= \int f \mathrm d \mu.
\end{align}
$$
Let's prove the reverse direction. Fix a continuous bounded $f:X \to \mathbb R$. Clearly, $f$ and $-f$ are l.s.c. and bounded from below. So
$$
\liminf_n \int f \mathrm d \mu_n \ge \int f \mathrm d \mu \quad \text{and} \quad \liminf_n \int -f \mathrm d \mu_n \ge \int -f \mathrm d \mu.
$$
Then
$$
\liminf_n \int f \mathrm d \mu_n \ge \int f \mathrm d \mu \quad \text{and} \quad -\limsup_n \int f \mathrm d \mu_n \ge -\int f \mathrm d \mu.
$$
This completes the proof.
