A chordally locally uniform meromorphic limit is meromorphic or identically infinity

A local uniform limit of continuous maps into the metric space $(\widehat{\mathbb{C}},\chi )$ is continuous, so $f$ is chordally continuous. If $f(a)\in \mathbb{C}$, choose a chordal neighbourhood $U$ of $f(a)$ whose closure misses $\mathrm{\infty }$. On $U$ the chordal and Euclidean metrics are comparable, and continuity of $f$ together with chordal local uniform convergence gives a neighbourhood $V$ of $a$ with $f(V)\cup f_n(V)\subseteq U$ for all large $n$. On $V$ the maps $f_n$ have no poles, hence are holomorphic there, and [L1] makes the Euclidean local uniform limit $f$ holomorphic near $a$.

If $f(a)=\mathrm{\infty }$, choose a chordal neighbourhood $U$ of $\mathrm{\infty }$ whose complement is a closed Euclidean disc. Continuity of $f$ and chordal local uniform convergence give a neighbourhood $V$ of $a$ with $f(V)\cup f_n(V)\subseteq U$ for all large $n$. The infinity-chart expressions $g=1\mathrm{/}f$ and $g_n=1\mathrm{/}{f}_n$ are then well-defined holomorphic maps on $V$, and the same metric comparison turns $g_n\to g$ into Euclidean local uniform convergence. Fact [L1] makes $g$ holomorphic, so $f$ is meromorphic at $a$.

When every $f_n$ is holomorphic, the functions $g_n$ of step 1.2 are holomorphic and nowhere zero on $V$. By [L2], their limit $g$ is either identically $0$ on $V$ or nowhere zero. Because $g(a)=0$, one gets $g\equiv 0$ on $V$, so $f\equiv \mathrm{\infty }$ on $V$. Thus the $\mathrm{\infty }$-value set of $f$ is open in the holomorphic-input case.

If $f$ takes some finite value, then steps 1.1 and 1.2 show that it is meromorphic at every point of $\mathrm{\Omega }$; otherwise $f\equiv \mathrm{\infty }$. In the holomorphic-input case, the finite-value set is open by step 1.1 and the $\mathrm{\infty }$-value set is open by step 2.1, so connectedness leaves only the two possibilities: $f$ is holomorphic on all of $\mathrm{\Omega }$, or $f\equiv \mathrm{\infty }$.