 Solution

Suppose a subsequence had $\mathrm{Im}{g}_A(z_n)\ge ϵ>0$. 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 givecontradicting $z\in A$. Thus $\mathrm{Im}{g}_A(z_n)\to 0$.

Convergence of the real parts can fail. For the vertical slitthe 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.