Let $\Omega\subset\mathbb{C}^n$ be holomorphically convex, and let $\Delta\subset\mathbb{C}$ be the unit disc. Let $\phi_i:\bar{\Delta}\to\Omega$ be continuous maps with $\phi_i:\Delta\to\Omega$ holomorphic. Prove that if $\cup_{i=1}^{\infty}\phi_i(\partial\Delta)$ is compact and is a subset of $\Omega$, then $\cup_{i=1}^{\infty}\phi_i(\Delta)$ is compact and is a subset of $\Omega$.
My idea: Let $K=\cup_{i=1}^{\infty}\phi_i(\partial\Delta)$. Since $K$ is compact, $K$ is closed. Note that $\Omega$ is holomorphically convex, $\hat{K}:=\{z\in\Omega:|f(z)|\le \sup_K|f|,\forall f,f$ is holomorphic$\}$ is compact and is a subset of $\Omega$. By maximum modulus principle, we know $\phi_i(\Delta)\subset \hat{K}$, hence $\cup_{i=1}^{\infty}\phi_i(\Delta)\subset\hat{K}\subset\Omega$. It suffices to show $\cup_{i=1}^{\infty}\phi_i(\Delta)$ is compact.
Any ideas?
This is called continuity principle (In fact, it is equivalent to holomorphic convexity). By maximum modulus principle, we know $\phi_i(\overline{\Delta})\subset \widehat{\phi_i(\partial\Delta)}.$ Let $d$ be the Euclidean distance, then we have $$d(\phi_i(\overline{\Delta}),\partial\Omega)\geq d(\widehat{\phi_i(\partial\Delta)},\partial\Omega)=d(\phi_i(\partial\Delta),\partial\Omega),$$ where the last equality follows from Thm 2.5.5 of Hormander. By assumption, there exists $\delta>0$ such that $$d(\phi_i(\partial\Delta),\partial\Omega)\geq \delta$$ for any $i,$ so $$d(\phi_i(\overline{\Delta}),\partial\Omega)\geq \delta$$ for any $i.$ Therefore, $$\mathop{\cup}_i\phi_i(\overline{\Delta})\subset\subset \Omega.$$