= Solution
Suppose a subsequence had $\operatorname{Im}g_A(z_n)\geq\epsilon>0$. The images cannot tend to infinity, because the inverse mapping-out function satisfies $g_A^{-1}(w)=w+o(1)$ at infinity while $z_n$ remains bounded. A further subsequence therefore converges to some $w\in\mathbb H$. Continuity of $g_A^{-1}$ inside $\mathbb H$ would give
$$
z=g_A^{-1}(w)\in\mathbb H\setminus A,
$$
contradicting $z\in A$. Thus $\operatorname{Im}g_A(z_n)\to0$.
Convergence of the real parts can fail. For the vertical slit
$$
A=\{iy:0<y\leq1\},
\qquad
g_A(z)=\sqrt{z^2+1},
$$
the two sides of the slit at $z=iy$, $0<y<1$, map to the two boundary values $\pm\sqrt{1-y^2}$. Alternating sequences approaching $iy$ from the two sides make $g_A(z_n)$ alternate between neighborhoods of these distinct limits.
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