= Solution
A <normal family> of <holomorphic functions> on a domain is one from which every sequence has a subsequence converging locally uniformly to a <holomorphic function>, or locally uniformly to infinity in the extended convention. For a bounded family, only finite limits occur.
For functions mapping a plane domain $D$ into the <unit disc>, the <Cauchy integral formula> supplies uniform derivative bounds on each compact subset. More explicitly, if a closed disc of radius $r$ around $z$ lies in $D$, the <Cauchy estimate> gives $|f'(z)|\leq1/r$. A slightly larger compact neighborhood of any given compact set therefore gives uniform boundedness and equicontinuity there. Apply the <Arzelà-Ascoli theorem> on a compact exhaustion and take a diagonal subsequence. Its limit is locally uniform; the <Cauchy integral formula> shows that the limit is <holomorphic>. This proves normality rather than merely invoking it.
Now let $|g|\leq M$ on the <complex upper half-plane>. Every dilation $g_t(w)=g(tw)$ has the same bound, so the argument just given, after scaling the bound if necessary, proves that $\{g_t:t\geq1\}$ is a <normal family>. For any sequence $t_j\to\infty$, every locally uniform subsequential limit $G$ satisfies
$$
G(iy)=\lim_jg(it_jy)=\ell\qquad(y>0).
$$
The <identity theorem for holomorphic functions> forces $G\equiv\ell$. It follows that $g_t\to\ell$ locally uniformly as $t\to\infty$: otherwise a sequence violating convergence on one compact set would have a convergent subsequence with this same constant limit, a contradiction.
For $0<\varepsilon<\pi/2$, the arc $K_\varepsilon=\{e^{i\theta}:\varepsilon\leq\theta\leq\pi-\varepsilon\}$ is a compact subset of the <complex upper half-plane>. Write $z=|z|w$ with $w\in K_\varepsilon$. Local uniform convergence of the dilations gives
$$
\boxed{g(z)\longrightarrow\ell\quad\text{as }|z|\to\infty\text{ in }S(\varepsilon).}
$$
For $\varepsilon\geq\pi/2$ the specified sector is empty. This is the same compactness mechanism as the <normal-family proof of angular boundary convergence>.
To transfer the result to the <unit disc>, use the <Cayley transform between the half-plane and disk>
$$
w=C_\omega(z)=i\frac{1+\overline\omega z}{1-\overline\omega z},\qquad
g(w)=h\left(\omega\frac{w-i}{w+i}\right).
$$
It sends $r\omega$ to $iy$ with $y=(1+r)/(1-r)$, so $g(iy)\to\ell$ as $y\to\infty$. Choose the curvature-minus-one normalization of the <hyperbolic metric>, for which the disc line element is $2|dz|/(1-|z|^2)$ and the half-plane line element is $|dw|/\operatorname{Im}w$. The <Cayley transform between the half-plane and disk> preserves these metrics. For $w\in\mathbb H$ and $y>0$,
$$
\cosh\rho_{\mathbb H}(w,iy)
=\frac{|w|^2+y^2}{2y\operatorname{Im}w}
\geq\frac{|w|}{\operatorname{Im}w}.
$$
Thus $\rho_{\mathbb D}(z,r\omega)<k$ implies $\operatorname{Im}w/|w|>\operatorname{sech}k$. This is the <hyperbolic tube around a radial segment> estimate. It places $w$ in a fixed sector bounded away from the real axis. Moreover $z\to\omega$ implies $|C_\omega(z)|\to\infty$. Applying the sector result gives
$$
\boxed{h(z)\longrightarrow\ell\quad\text{as }z\to\omega\text{ within }\Sigma(k).}
$$
A different constant normalization of the <hyperbolic metric> only rescales $k$ and gives the same conclusion.
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