Most of what occurs to me has already been said, but you may find the following picture useful.
If $d_C$ is the Chebyshev metric on $R^2$, i.e. with points $\mathbf{p} = (x_1,y_1)$ and $\mathbf{q} = (x_2,y_2)$ in $R^2$,
$d_C(\mathbf{p,q}) := |x_1-x_2| \vee |y_1-y_2|$,
and $h_C$ is the Hausdorff metric on closed subsets of $R^2$ induced by $d_C$, i.e. with $A$ and $B$ being closed subsets of $R^2$,
$h_C(A,B):= \sup_{\mathbf{p} \in A} d_C(\mathbf{p},B) \vee \sup_{\mathbf{q} \in B} d_C(\mathbf{q},A)$,
where as usual $d_C(\mathbf{p},B) = \inf_{\mathbf{r} \in B} d_C(\mathbf{p,r})$ etc,
then the Levy metric between two distribution functions $F$ and $G$ is simply the Hausdorff distance $d_C$ between the closures of the completed graphs of $F$ and $G$.
Set
$$K(F,G) := \inf\{\varepsilon>0; G(x-\varepsilon) \leq F(x) \leq G(x+\varepsilon) \, \text{for all} \, x \in \mathbb{R}\}.$$
It is not difficult to see that $K$ is a metric. Let us consider the following easy example in order to see the differences between the Lévy metric $L$ and the metric $K$:
Define random variables $X:= 1_{[1/2,1]}$ and $X_n := \frac{1}{2} 1_{(0,1/n]}+1_{[1/2,1]}$ on the unit interval $[0,1]$ (endowed with the Lebesgue measure). The corresponding distribution functions are given by
$$G(x) := \mathbb{P}(X \leq x) = 1_{[1,\infty)}(x) \qquad \qquad G_n(x) := \mathbb{P}(X_n \leq x) = \frac{1}{n} 1_{[1/2,1)}(x) + 1_{[1,\infty)}(x).$$
Using the definition of $L$ and $K$ it is not difficult to see that $L(G_n,G) \to 0$ whereas $K(G_n,G)=1/2$. On the other hand, we know that $X_n \to X$ almost surely; that's why the Lévy metric $L$ is more natural: If the random variables converge almost surely, then we would expect that the distance of the corresponding distribution functions converges to $0$.
A typical $\varepsilon$-neighborhood of the distribution function $G$ with respect to the Lévy metric $L$ looks like that
whereas a neighborhood with respect to $K$ is of the following form
Remark In fact, one can show that $L(G_n,G) \to 0$ if and only if $X_n \to X$ in distribution for any random variables $(X_n)_n$ and $X$.
Best Answer
"$\Leftarrow$": Suppose that $d_L(\mu_n,\mu) \to 0$ as $n \to \infty$. Fix a sequence $(\varrho_n)_{n \in \mathbb{N}}$ such that $\varrho_n \to 0$ as $n \to \infty$ and $$d_L(\mu_n,\mu) = d_L(\mu,\mu_n) < \varrho_n.$$
By the very definition of the Lévy metric, there exists $\epsilon_n \in [0,\varrho_n]$ such that
$$F_{\mu}(x-\epsilon_n)-\epsilon_n \leq F_{\mu_n}(x) \leq F_{\mu}(x+\epsilon_n) + \epsilon_n, \qquad x \in \mathbb{R}. \tag{1} $$ Note that $\epsilon_n \to 0$ as $n \to \infty$. Now if $x$ is a continuity point of $F_{\mu}$, then
$$F_{\mu}(x-\epsilon_n)-\epsilon_n \to F_{\mu}(x) \quad \text{and} \quad F_{\mu}(x+\epsilon_n)+\epsilon_n \to F_{\mu}(x),$$
and therefore it follows from $(1)$ that $F_{\mu_n}(x) \to F_{\mu}(x)$ as $n \to \infty$.
"$\Rightarrow$": Suppose that $F_{\mu_n}(x) \to F_{\mu}(x)$ for all continuity points of $F_{\mu}$. Fix $\epsilon>0$. Since $F_{\mu}$ is monotone and bounded, it can have at most countably many jumps, and so it is continuous up to a countable set $D$ of discontinuity points. In particular, we can find continuity points $x_1 < \ldots < x_k$ such that $x_{i+1}-x_i < \epsilon$ for all $i$ and
$$F_{\mu}(x_1) < \frac{\epsilon}{2} \qquad \text{and} \qquad F_{\mu}(x_k)>1-\frac{\epsilon}{2}. \tag{2}$$
Choose $N \in \mathbb{N}$ sufficiently large such that
$$|F_{\mu_n}(x_i)-F_{\mu}(x_i)| \leq \frac{\epsilon}{2} \qquad \text{for all $n \geq N$, $i=1,\ldots,k$}. \tag{3}$$
For any $x \in [x_{i-1},x_i]$, we have by the monotonicity of $F_{\mu}$, $F_{\mu_n}$ and $(3)$
$$F_{\mu_n}(x) \leq F_{\mu_n}(x_i) < F_{\mu_n}(x_i)+ \frac{\epsilon}{2} + F_{\mu}(x_i)-F_{\mu}(x_i) \stackrel{(3)}{\leq} F_{\mu}(x_i)+\epsilon \leq F_{\mu}(x+\epsilon)+\epsilon$$
for all $n \geq N$. For $x<x_1$ it holds that
$$F_{\mu_n}(x) \leq F_{\mu_n}(x_1)-F_{\mu}(x_1)+F_{\mu}(x_1) \stackrel{(2),(3)}{<} \frac{\epsilon}{2} + \frac{\epsilon}{2} \leq F_{\mu}(x+\epsilon)+\epsilon$$
and for $x>x_k$
$$F_{\mu_n}(x) \leq 1 \stackrel{(2)}{\leq} F_{\mu}(x+\epsilon)+\epsilon.$$
This proves $F_{\mu_n}(x) \leq F_{\mu}(x+\epsilon)+\epsilon$ for all $x \in \mathbb{R}$ and $n \geq N$. A very similar reasoning shows
$$F_{\mu}(x-\epsilon)-\epsilon \leq F_{\mu_n}(x),$$
and since $\epsilon>0$ is arbitrary, this finishes the proof.