I've read and understood the proof for the complex plane for the analytic continuation theorem. The proof relies on $E$ being both closed and open, with $\quad E = \bigcap _ { n \geq 0 } E _ { n }$ and $E _ { n } = \{ z \in \Omega | f ^ { ( n ) } ( z ) = 0 \}$, so it can only be $\varnothing$ or $\Omega$ (the open set where the holomorphic function is defined).
While I get this, my intuition (I'm a physicist) of the analytic continuation theorem was that functions couldn't be zero on open subsets because the derivatives of $f$ on $x$ and $y$ are connected. I don't know how to connect this with the proof above.
Related to this question, does analytic continuation work on $\mathbb{R}^1$ functions? what about $\mathbb{R}^2$ functions? It should work on $\mathbb{R}^2$ because it's isomorphic to $\mathbb{C}$ but I might be wrong.
This questions might be super basic to most mathematicians but as a physicist I struggle quite a bit. Maybe because the're so basic I didn't found a lot on the internet.
Thanks!