I understand that an iterative argument can be used here, where one considers the points in a neighbourhood of the point at which the function is zero. However, I do not know how to go about proving this exactly.
In summary, I am trying to prove the strong unique continuation property of real analytic functions (in a topological space).
Let $I$ be an open interval in $\mathbb R$, let $J$ be another open interval, such that $J\supset I$, and let $f,g\colon J\longrightarrow\mathbb R$ be two analytic functions such that $f|_I=g|_I$. You want a proof of the fact that $f=g$, I suppose.
If $I=(a,b)$ and $b\in J$, then consider the power series $\sum_{n=0}^\infty a_n(x-b)^n$ and $\sum_{n=0}^\infty b_n(x-b)^n$ that represent $f$ and $g$ respectively near $b$. Then from the fact that $f=g$ to the left of $b$, you can deduce that $(\forall n\in\mathbb{Z}^+):a_n=b_n$ (since $a_n=\frac{f^{(n)}(b)}{n!}=\frac{g^{(n)}(b)}{n!}=b_n$). But then the equality $f(x)=g(x)$ is true in some interval $(a,b+\varepsilon)$. By the same argument, if $a\in J$, then the equality $f(x)=g(x)$ is true in some interval $(a-\varepsilon,b)$. So, you can always enlarge the interval $I$ and therefore $I=J$.