= Solution
<Morera's theorem> states that a continuous complex-valued function on an open set is <holomorphic> if its integral around every triangle whose closed interior lies in the set is zero. It is enough to impose this condition locally on disks.
Fix a <disk> $B$ contained in the open set and a base point $z_0\in B$. Define $F(z)$ as the integral of $f$ along the straight segment from $z_0$ to $z$. The zero triangle integral gives, for sufficiently small $h$,
$$
F(z+h)-F(z)=\int_{[z,z+h]}f(w)\,dw
=h\int_0^1 f(z+th)\,dt.
$$
<Continuity> makes the quotient by $h$ tend to $f(z)$, so $F$ is <holomorphic> and $F'=f$. To justify that its derivative is <holomorphic>, apply the <Cauchy integral formula> for $F$ on a smaller <circle>:
$$
F'(z)=\frac1{2\pi i}\int_{\Gamma}\frac{F(w)}{(w-z)^2}\,dw.
$$
The right side may be differentiated again inside the <circle> because its denominator stays uniformly away from zero on the <contour>. Thus $F'=f$ is <holomorphic> on $B$. Such disks cover the open set, proving the theorem.
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