Let $\lbrace X_n \rbrace_{n \in \mathbb{N}}, X$ be random variables, and let $\lbrace F_n \rbrace_{n\in \mathbb{N}}, F$ be their respective distribution functions. Assume that $X_n \xrightarrow{d}X$, and that $F$ is continuous. Prove that, for $a <b$, \begin{align*} \sup_{x \in [a,b]}|F_n(x) - F(x)| \to 0, \quad n\to \infty \end{align*}
So my professor gave me a hint for this problem (However, I'm not able to complete it yet). The idea is to prove that for any sequence $x_m$ such that $x_m \to x_0$ as $m\to \infty$ $F_m(x_n) \to F(x_0)$ as $m,n \to \infty$. Now, prove that $f_n$ converge to $f$ uniformly on $[a,b]$. So this is my attempt.
First, we assume that $\lbrace x_m \rbrace_{m\ in \mathbb{N}}$ is a sequence such that $x_m \to x_0$ as $m \to \infty$. Since $F$ is continuous, then $F(x_m) \to F(x_0)$ as $m \to \infty$.
Moreover, for all $m \in \mathbb{N}$, since $F$ is continuous, then $x_m$ is the continuity point of $F$. Since $X_n \xrightarrow{d} X$, then $F_n(x_m) \to F(x_m)$ as $n \to \infty$. Therefore, for all $m,n \in \mathbb{N}$, \begin{align*} |F_n(x_m)-F(x_0)| = |F_n(x_m)-F(x_m)+F(x_m)-F(x_0)| \leq |F_n(x_m)-F(x_m)| + |F(x_m)-F(x_0)|. \end{align*} Let $n\to \infty$, $|F_n(x_m) -F(x_m)| \to 0$ for all $m\in \mathbb{N}$. Next, let $m\to \infty$, $|F(x_m) -F(x_0)| \to 0$. Therefore, we conclude \begin{align*} F_n(x_m)\to F(x_0), \end{align*} as $n,m \to \infty$.
Now, we recall that for a sequence of function on $[a,b]$, say $\lbrace f_n\rbrace_{n\in \mathbb{N}}$ on $[a,b]$, and $f$ on $[a,b]$. Then $f_n \to f$ uniformly if and only if $\sup\limits_{x\in [a,b]}\lbrace|f_n(x)-f(x)| \rbrace \to 0$ as $n\to \infty$. So we now want to prove that $F_n \to F$ uniformly on $[a,b]$.
Assume not, then there exists $\epsilon >0$ such that for all $n \in \mathbb{N}$, there exists $m\in \mathbb{N}, m\leq n$, there exists $x \in [a,b]$ such that \begin{align*} |F_m(x)-F(x)| \geq \epsilon. \end{align*} Imply for $n=1$, there exists $n_1 \in \mathbb{N}, n_1 \geq 1$ and $x_1 \in [a,b]$ such that $|F_{n_1}(x_1)-F(x_1)| \geq \epsilon$.
By induction, we construct a sequence $\lbrace x_k \rbrace_{k \in \mathbb{N}} \subset [a,b]$ such that \begin{align*} |F_{n_k}(x_k)-F(x_k)| \geq \epsilon \quad \forall k\in\mathbb{N}. \end{align*} Since $n_k \geq k$ for all $k \in \mathbb{N}$, then $\lbrace F_{n_k}(x)\rbrace$ is a subsequence of $\lbrace F_k (x) \rbrace$ for all $x\in [a,b]$. Then $F_{n_k}(x_k) \to F(x_k)$ as $n_k \to \infty$ for all $x\in [a,b]$.
So, my problem is that I cannot show that $\lbrace x_k \rbrace$ is convergent. If I had such thing, I can show the contraction by using Bolzano-Weierstrass. Can somebody check the proof for me and give any hints if possible?