let $F_n \implies F$ where the symbol $\implies $ means pointwise convergence for every x that is a point of continuity of F.
Suppose now that F is continuous. Show that $F_n$ converges uniformly to $F$, that is
$$sup_x |F_n (x)-F (x)|\to 0, n\to\infty$$
(That's exercise 5. Chapter 3 paragraph 1 of Albert Shiryaev's book Probability)
Now, Convergence in generality with F continuous basically means pointwise convergence for all x, but that doesn't imply uniform convergence. Is there any assumptions that may be missing? Or does this really imply uniform convergence for some reason?
Sorry for the lack of development, usually we can use compactness to show uniform convergence but that wasn't part of the assumptions.
Hint
Usually when things that converge pointwise fail to converge uniformly, the "part of the function that fails to get close" is sent to infinity or squeezed onto the boundary of the region as $n\to \infty$. For instance, $f_n(x) = e^{-(x-n)^2}$ is a lump of the same height for all $n$ and thus never gets uniformly close to the zero function, but the lump gets sent off to infinity as $n\to\infty$, eventually passing every value of $x,$ allowing each point to eventually get close to zero.
Can you see why, given Sangchul's comment, this kind of behavior wont be possible when $F_n$ and $F$ are distribution functions?