Families omitting two values are chordally normal

If $\mathcal{F}$ is empty, it is chordally normal vacuously. Otherwise fix $a\in \mathrm{\Omega }$ and choose $\rho >0$ with $\overline{D(a,2\rho )}\subset \mathrm{\Omega }$. For each $f\in \mathcal{F}$, at least one of $\mathrm{\mid }f(a)\mathrm{\mid }$, $\mathrm{\mid }1\mathrm{/}f(a)\mathrm{\mid }$, or $\mathrm{\mid }1\mathrm{/}(1-f(a))\mathrm{\mid }$ is at most $2$: if both $\mathrm{\mid }f(a)\mathrm{\mid }<1\mathrm{/}2$ and $\mathrm{\mid }1-f(a)\mathrm{\mid }<1\mathrm{/}2$ held, the triangle inequality would fail. Thus one of the three transforms $T_1(z)=z$, $T_2(z)=1\mathrm{/}z$, $T_3(z)=1\mathrm{/}(1-z)$ has center value of modulus at most $2$.

Each $T_j\circ f$ omits $0$ and $1$, so [L1] applied after rescaling $D(a,2\rho )$ to $\mathbb{D}$ gives a bound $\mathrm{\mid }{T}_j(f(z))\mathrm{\mid }\le M(\rho )$ on $D(a,\rho )$ for the transform selected in step 1.1. Hence every member of the transformed family is locally bounded there. Fact [L2] makes each transformed subfamily locally equicontinuous, and because there are only three fixed inverse transforms, the original family is chordally locally equicontinuous on $D(a,\rho )$.

The target $\widehat{\mathbb{C}}$ is compact, so pointwise relative compactness is automatic. Therefore [A1] and [L3] apply on each $D(a,\rho )$, giving chordal normality there. As $a$ was arbitrary, $\mathcal{F}$ is chordally normal on $\mathrm{\Omega }$.