Here is the theorem I have been attempting to prove, I would love to hear some feedback on it!
Theorem:
Suppose $\Omega$ is a connected open subset of $\mathbb{R}^{n}$, $f$ is a real analytic function on $\Omega$ and $f=0$ on a non empty open subset of $\Omega$. Then $f\equiv 0$ in $\Omega$.
Proof:
Let $\omega$ be the interior (definition 2.3.4) of $\{x\in\Omega : f(x)=0\}$. Then $\omega$ is open, and by the assumption, it is non-empty. Let $a\in\Omega$ be a limit point of $\omega$ (definition 2.3.5). As $f$ is real analytic everywhere in $\Omega$, it is real analytic at the limit point $a\in \Omega$. Meaning that the power series expansion of $f$ about the point $a$ is convergent in some neighbourhood of $a$. Now, any neighbourhood of $a$ must intersect $\omega$, and $f$ is identically zero in $\omega$. Now, all derivatives vanish of infinite order in the interior of $\omega$, so by continuity all derivatives must vanish of infinite order at the point $a$. Therefore, it must be that $f=0$ on a neighbourhood of $a$, which implies that $a\in \omega$. Since $\omega$ contains all of its limit points in $\Omega$, it is relatively closed. However, it was stated earlier that $\omega$ is an open set of $\Omega$. Hence, $\omega$ is clopen in $\Omega$.
Since $\omega$ is clopen in the connected space $\Omega$, and it is non-empty, it must be that $\omega=\Omega$ (by proposition 2.3.1).