Every harmonic function is the real part of a holomorphic function

Is the following "integral free" enough?

Let $$h:\quad \Omega\to{\mathbb R}\ ,\qquad (x,y)\mapsto h(x,y)$$ be the given harmonic function which we assume to be $C^2$. Using $h$ we define the function $$f(x,y):= h_x(x,y)- i\ h_y(x,y)$$ with real part $u(x,y)=h_x(x,y)$ and imaginary part $v(x,y)=- h_y(x,y)$. As $h$ is harmonic one has $u_x\equiv v_y$, furthermore $u_y\equiv -v_x$ by equality of the mixed derivatives. So $f\colon \Omega\to{\mathbb C}$ is $C^1$ and satisfies the CR equations; therefore it is an analytic function of $z=x+iy\in\Omega$.

Assume that $\Omega$ is simply connected and chose a point $z_0\in\Omega$. Then by a standard theorem of complex analysis the function $$F(z)\ :=\ h(z_0)+\int_\gamma f(z)\ dz\ , \qquad \hbox{$\gamma\ $ a path from $z_0$ to $z$}\ ,$$ is an analytic primitive of $f$ in $\Omega$. Let $(x,y)\mapsto U(x,y)$ be the real part of $F$. Then by the CR equations, this time applied to $F$, we have $$U_x(z) -i U_y(z)=F'(z)=f(z)= h_x(z)-i h_y(z)\qquad(z\in\Omega)\ .$$ It follows that $$\nabla U(x,y)\equiv\nabla h(x,y)\qquad \bigl( (x,y)\in\Omega\bigr)\ ,$$ and as $U(x_0,y_0)=h(x_0,y_0)$ we conclude that in fact $U(x,y)\equiv h(x,y)$ in $\Omega$.

Note that we had to assume $\Omega$ simply connected. The function $h(x,y):=\log\sqrt{x^2+y^2}$ is harmonic in the punctured plane but is not the real part of an analytic function in this domain.

The statement is false without assumptions on the domain; log$| z |$ is harmonic on the punctured plane and can locally be expressed as the real part of an analytic function, but can't be the real part of an analytic function on any annulus about $0$ since the argument can't even be continuous.

As an aside, "annulus" gets spell-checked on this site! haha