Proof verification: If $f$ is holomorphic on an open set and if $Re(f)$ is constant, then $f$ is constant.

Question

Suppose $Re(f)$ is constant. I am trying to prove that $f$ itself is constant. To do so, I want to show that for every $z_0$in our open set, $f'(z_0) = 0.$ We know that $f'(z_0) = \frac{\partial f}{\partial x}(z_0).$ This is where I make an iffy step. We can write $f$ as $f(x, y) = u(x, y) + v(x, y)i.$ Then $\frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial x}i.$ By the Cauchy formulas, we have $\frac{\partial f}{\partial x} = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}i.$ However, since $Re(f)$ is constant, we know that $\frac{\partial u}{\partial x} = \frac{\partial u}{\partial y} = 0.$ Hence, $\frac{\partial f}{\partial x}(z_0) = 0.$ Does this work?