= Solution
First take $z\in U_-$ and put $w=\bar z\in U_+$. Reflection symmetry of the function gives
$$
\frac{f(z+h)-f(z)}h
=\overline{\frac{f(w+\bar h)-f(w)}{\bar h}}
\longrightarrow\overline{f'(w)}.
$$
Hence $f$ is <holomorphic> on $U_-$, with $f'(z)=\overline{f'(\bar z)}$ there.
It remains to cross the real axis. Fix a real point $p\in U$ and a small disk about $p$ with closure in $U$. Consider a <triangle> in a still smaller disk. Intersect its filled interior with the closed upper and lower half-planes and triangulate the resulting <convex> polygons. This gives finitely many <triangles> lying on one side of the real axis, possibly with edges on that axis. Degenerate pieces contribute zero.
For an upper piece, translate its three vertices by $i\varepsilon$ with $\varepsilon>0$. For sufficiently small $\varepsilon$ the translated <triangle> lies entirely in $U_+$ and remains in $U$, so its boundary <integral> is zero by Cauchy's theorem. Part (d), or its segment-integral proof, shows that the <integral> converges to that around the untranslated piece as $\varepsilon\downarrow0$. Thus the original piece has zero <integral>. Translate lower pieces by $-i\varepsilon$ to obtain the same conclusion in $U_-$.
When all the oriented <triangle> integrals are added, their internal edges cancel, including the segments on the real axis. The original <triangle> therefore has zero boundary <integral>. <Continuity> of $f$ on $U$ now allows <Morera's theorem> in the disk, proving holomorphicity near $p$. Together with the two half-plane arguments this gives
$$
\boxed{f\text{ is holomorphic throughout }U.}
$$
This is a <Morera gluing lemma across a straight line>; the original <continuity> hypothesis is essential to the limiting step.
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