$U$ a harmonic function, prove if $U(z)=0$ on some ball in a region $G$, then $U \equiv 0$ on $G$ [duplicate]

The question: Show that if $u$ is real and harmonic on a connected open set $U \subseteq \mathbb{C}$, and $ u = 0$ on some small disc $D(P,r) \subseteq U$, then $u = 0$ on all of U.

Now here is a couple of ways I thought of solving this problem: The first way is this: since $u$ is harmonic, there exists a holomorphic function $F$, on set $U$ such that $Re(F) = u$ on $U$. In my textbook, there is one corollary which says that if a holomorphic function $F = 0$ on some disc $D(P,r)$ which is a subset of a connected open set $U$ then $F = 0$ on $U$. The problem is that since $F$ is holomorphic such that $Re(F) = u = 0$ on the disc $D(P,r)$, $F$ is constant on $U$ but not exactly zero.

My second option is to use the mean value property on the disc $D(P,r)$. We have $u(P) = \int_{0}^{2 \pi} u(P + re^{i \theta}) d\theta = 0$, since $u = 0$ on the disc. My idea is the construct a disc $D(P, r_{0})$ such that the disc is a subset of U and the boundary of U. Do you think I am on the right track? Thank you for your help!

Let A the set of $z_0 \in U$ for which $u$ is zero in a neighborhood of $z_0$. A is obviously open, but now if $z_1 \in U$ is in the closure of A (in U), let B a small open disc containing $z_1$ and included in U; then there is an $F_B$ analytic on B, $\Re{F_B}=u$ on B; but there is a $z_0$ in A also contained in B (as $z_1$ is in the closure of A!), so there is a small open set C included in B on which $u=0$, hence $F_B$ is purely imaginary on C, hence constant on C (by the open mapping property for example), hence constant on B by the identity principle, hence $z_1$ has the property that defines A, so it is in A. This means A is open and closed in U, so being non-empty by hypothesis, it is all U, done!