The theorem states that if $f$ and $g$ are analytic functions and their values agree on an open set that is contained in a larger, connected domain, then $f$ must equal $g$ on the entire domain. (The domain need not be simply connected, which is the part that confuses me most.)
Instead of the set-theoretic proof (the clopen set proof), can we understand it from the point of view of power series expansions?
Let $f = g$ on an some open set $S$, which is contained in larger connected domain $\Omega$. $f$ and $g$ are analytic on this domain.
Then look at the function $f-g = 0$ on this open set. Its Taylor expansion is identically zero.
Can I now use this argument for the whole domain? And this is what is challenging - there can be a hole in the domain. So, I can't argue "$f-g = 0$ converges in the largest Taylor disk possible, so its Taylor series is identically zero everywhere in the domain; therefore, $f=g$ everywhere on the domain".
EDIT: I'm sketching some small, overlapping disks of convergence of $f-g = 0$, on a connected domain with a hole in it. If I start at the first known disk of convergence, the disk that overlaps with the first disk must also have all zero terms in its Taylor expansion, i.e., $f-g$ is again identically zero, but we've extended the region in the domain where we know the values of $f-g = 0$. And iterate this process, until we have enough overlapping disks to cover the connected domain - all disks will have Taylor expansion identically zero.