Finding the zero set of the gradient of a harmonic function

Question

I am trying to find the vanishing set of the gradient of a harmonic function $u$ defined in a region $\Omega$. I have tried the real case (I have verified that once the real case is shown, the complex case follows easily) but am getting stuck. This is how I have tried.
We first consider the case where $u$ is real-valued. Then for every $z\in\Omega$ there is an open disc $B(z;r)$ centred at $z$ in $\Omega$, such that $u$ is the real part of a holomorphic function $f=u+iv$ defined on $B(z;r)$.
Since $f$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann equations \begin{align*} \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\;;\;\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} \end{align*} which tell us that the zero set of the gradient $(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y})$ of $u$ in $B(z;r)$ and that of the gradient $(\frac{\partial v}{\partial x},\frac{\partial v}{\partial y})$ of $v$ are equal. Let us call this set $A$. Then for each $z\in A$, $f'(z)=\frac{\partial u}{\partial x}(z)+i\frac{\partial v}{\partial x}(z)=0+i0=0$. Thus $A$ is the set of zeroes of $f'$ as well.
Now, $f'$ is the derivative of a holomorphic function, and thus is holomorphic itself. We know that the zeros of a non-constant holomorphic function on a region form a discrete closed subset of that region. So either $f'\equiv 0$, or $A$ is a discrete closed subset of $B(z;r)$. If $f'\equiv 0$, this means the gradient of $u$ vanishes on an open disc in $\Omega$, and hence the harmonic function $u$ is constant on an open disc in $\Omega$, and thus by the identity theorem for harmonic functions, on entire $\Omega$ ($\Omega$ being a region). Thus the gradient of $u$ vanishes everywhere on $\Omega$. I am however not being able to proceed with the case where $A$ is a discrete closed subset of $B(z;r)$. Any help will be appreciated.