Weak convergence in probability implies uniform convergence in distribution functions

Fix $\epsilon>0$. Since $F$ is a distribution function, there exists $R >0$ such that $F(r) \leq \epsilon$ for all $r \leq -R$ and $F(r) \geq 1-\epsilon$ for all $r \geq R$. As $F_n \to F$ pointwise, we can choose $N \in \mathbb{N}$ such that

$$|F_n(-R) - F(-R)| \leq \epsilon \qquad \text{and} \qquad |F_n(R)-F(R)| \leq \epsilon$$

for all $n \geq N$. Hence, by the monotonicity of $F_n$ and $F$,

$$\begin{align*} |F_n(r)-F(r)| \leq |F_n(r)|+|F(r)| &\leq F_n(-R)+F(-R) \\ &= (F_n(-R)-F(-R)) + 2 F(-R) \\ &\leq 3\epsilon \tag{1} \end{align*}$$

for all $r \leq -R$. Similarly, it follows from

$$1 \geq F_n(r) \geq F_n(R) = (F_n(R)-F(R))+F(R) \geq 1-2\epsilon, \qquad r \geq R,$$

that

$$|F_n(r)-F(r)| \leq |F_n(r)-(1-\epsilon))|+ |F(r)-(1-\epsilon)| \leq 2 \epsilon \tag{2}$$

for all $r \geq R$. Combining $(1)$ and $(2)$ yields

$$\sup_{r \in [-R,R]^c} |F_n(r)-F(r)| \leq 3 \epsilon$$

for all $n \geq N$. Since you have already shown that $F_n$ converges to $F$ uniformly on compact intervals, there exists $N' \in \mathbb{N}$ such that

$$\sup_{r \in [-R,R]} |F_n(r)-F(r)| \leq \epsilon$$

for all $n \geq N'$. Setting $\tilde{N} := \max\{N,N'\}$, we get

$$\sup_{r \in \mathbb{R}} |F_n(r)-F(r)| \leq 3 \epsilon \qquad \text{for all $n \geq N$}.$$

Here's another proof for the compact case which I find a bit more intuitive. We are trying to show that $F_n \rightarrow F$ uniformly on an arbitrary closed interval $[a,b]$.

Fix $\epsilon > 0$. Let $d = F(b)-F(a)$, and take $k$ large enough so that $\frac{d}{k} \leq \frac{\epsilon}{5}$. By the continuity of $F$, we can apply the intermediate value theorem to show there exist real numbers $a := x_0 < x_1 < ... < x_k := b$ such that $F(x_i) = F(a) + i\frac{d}{k}$ for each $i \in \{0,1,...,k\}$.

Since $F_n \rightarrow F$ pointwise, for each $i$ in $\{0,1,...,k\}$ there exists $N_i$ such that $|F_n(x_i)-F(x_i)| \leq \frac{\epsilon}{5}$ for all $n \geq N_i$. Taking $N = \max(N_0,N_1,...,N_k)$, we conclude that $|F_n(x_i)-F(x_i)| \leq \frac{\epsilon}{5}$ for all $n \geq N$, $i \in \{0,1,...,k\}$. This gives $F_n(x_{i+1})-F_n(x_i) \leq |F_n(x_{i+1})-F(x_{i+1})| + |F(x_{i+1})-F(x_i)| + |F(x_i)-F_n(x_i)| \leq \frac{\epsilon}{5} + \frac{d}{k} + \frac{\epsilon}{5} \leq \frac{3\epsilon}{5}$

For any $x \in [a,b)$, we have $x_i \leq x < x_{i+1}$ for some $i \in \{0,1,...,k\}$. Then for all $n \geq N$, we have $$ \begin{align} |F_n(x)-F(x)| & \leq |F_n(x)-F_n(x_i)| + |F_n(x_i) - F(x_i)| + |F(x_i)-F(x)| \\\\ & \leq |F_n(x_{i+1})-F_n(x_i)| + |F_n(x_i)-F(x_i)| + |F(x_{i+1})-F(x)| \\\\ & \leq \frac{3\epsilon}{5} + \frac{\epsilon}{5} + \frac{d}{k} \\\\ & \leq \epsilon \\\\ \end{align} $$ where the second inequality uses monotonicity of $F$ and $F_n$.

Of course, $|F_n(b)-F(b)| \leq \frac{\epsilon}{5} \leq \epsilon$ since $b = x_k$.

Thus, we've shown $|F_n(x)-F(x)| \leq \frac{5\epsilon}{3}$ for all $x \in [a,b]$, concluding the proof.