Let be $S:=\sum\limits_k^{\infty}a_kx^k$ a real valued power series with convergence radius of $R>0$. Then, we know that the power series converges uniformly on every compact subset of $(-R,R)$.
Why do we require a compact set instead of an open set?
We could simply choose an $r$ with $0<r<R$ and then define an interval $M:=(-r,r)$. If I go through the proof which shows that $S$ is pointwise convergent on $(-R,R)$ I don't see where it should fail if I choose $r$ as an upper bound for all $x\in(-r,r)$? So I would conclude that $S$ is uniformly convergent on each open subset $(-r,r)$ with $0<r<R$ of $(-R,R)$.
We require a compact set because we have to exclude $(-R,R)$ itself. For any open subset you mention, the closure is then compact and also contained in $(-R,R)$, so the power series converges uniformly there, hence also on the open set.
Generally, compact convergence is quite common because as you mentioned power series converge compactly, a lot of other stuff also converges compactly (for example continuous functions on $\mathbb{R}^n$ converging pointwise to another continuous function). And they have the property that you can switch limits: if $x_n\to x$ and $f_n\to f$ compactly, then $$\lim_{n\to\infty}\lim_{m\to\infty} f_n(x_m) = \lim_{m\to\infty}\lim_{n\to\infty} f_n(x_m),$$ so that you can for example exchange integration or differentiation with compact limits.
In a more abstract setting, if $X$ is a locally compact topological space, the compact topology is the natural topology to give to $C(X,Y)$ (the space of continuous functions to another space $Y$), and the functor $C(X,-)$ is the right-adjoint for the functor $-\times X$ (topological tensor-hom adjunction).