= Solution
Fix $r<1$ and construct $g_n$ from the boundary values of $u_n$ on the circle $|z|=r$ as in <3/b/ii>. By the assumed <locally uniform convergence>, those boundary values converge uniformly to $u$. For each $s<r$, the kernel is bounded on $|z|\leq s$, so the same integral shows that $g_n$ converges uniformly there to
$$
g(z)=\frac1{2\pi}\int_0^{2\pi}u(re^{i\theta})\frac{re^{i\theta}+z}{re^{i\theta}-z}\,d\theta.
$$
The <Weierstrass convergence theorem> makes $g$ holomorphic. Since $\operatorname{Re}g_n=u_n$, passing to the limit gives $u=\operatorname{Re}g$ on $|z|<r$. The real part of a <holomorphic function> is harmonic. As $r<1$ was arbitrary, \b[$u$ is harmonic throughout the unit disc]. This proves the <locally uniform limit of harmonic functions> theorem without presuming convergence of the derivatives of $u_n$.
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