I am not seeing how to adapt my argument to produce a proof for the following statement:
Let $D \subset \mathbb{C}_{\infty}$ omit at least two points, and suppose that $\infty \notin \partial D$ and $\lim_{z \rightarrow \zeta_{0}} H_D \phi(z) = \phi(\zeta_{0})$ for every continuous function $\phi : \partial D \rightarrow \mathbb{R}$. Here $H_D\phi(z) = \sup_{u \in \mathcal{U}_{\phi}} u(z)$, where $\mathcal{U}_{\phi}$ is the family of all subharmonic functions $v$ defined on $D$, satisfying $\limsup_{z \rightarrow \zeta} v(z) \leq \phi(\zeta)$ for all $\zeta \in \partial D$, is the Perron solution to the Dirichlet problem in $D$, with boundary values $\phi$.
Then there exists a subharmonic barrier $b$ defined on an open set $N \cap D$, where $N$ is an open neighhbourhood of $\zeta_{0}$. That is, there exists a subharmonic function $b$ on $N \cap D$, where $N$ is an open neighbourhood of $\zeta$ in $\mathbb{C}_{\infty}$, satisfying $b < 0$, and $\lim_{z \rightarrow \zeta_{0}} b(z) = 0$.
$\textbf{My attempt}$: Consider $\phi(\zeta) = -|\zeta - \zeta_{0}|$ on $\partial D$. $H_D \phi$ is harmonic in $D$, and we necessarily have that $H_D\phi \leq 0$ on $D$, for otherwise (if $H_D \phi(z) > 0$ at some point $z \in D$), some $u \in \mathcal{U}_{\phi}$ would satisfy $u(z) > 0$, and $\limsup_{z \rightarrow \zeta} u(z) \leq 0$ at every $\zeta \in \partial D$, violating the maximum principle for subharmonic functions.
Thus $H_D\phi \leq 0$ on $D$. We would be able to show $H_D\phi$ is in fact a subharmonic barrier at $\zeta_0$ if we could show $H_D\phi < 0$ on an open set $N \cap D$, where $N$ is a sufficiently small open neighbourhood of $\zeta_0$. But I have tried to show this to no avail. I realize that if $H_D \phi$ were to equal zero at some point of $D$, it would be constant by the strong maximum principle. Now I fail to derive a contradiction based upon this conclusion.
Thank you in advance for any suggestions.
Here is a possible solution:
It is a fact that $H_{D} \phi \leq -H_{D} (-\phi) $ on $D$. $-\phi$ is itself subharmonic on $D$, so that $H_{D}(-\phi) \geq -\phi$ and $H_D \phi \leq -H_{D} (-\phi) \leq \phi$ on $D$, i.e. $H_D \phi (z) \leq -|z - \zeta_{0}|$ for all $z \in D$. But this means $H_D \phi$ is strictly negative on $D$, and is thus a barrier at $\zeta_{0}$ as required.