continuous extension of Banach space-valued analytic function

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Let $X$ be a Banach space and let $U\subseteq\mathbb{C}$ be a bounded open subset of the complex plane. Suppose that $f:U\to X$ is an analytic function.

It is well-known that $f$ may not admit a continuous extension to the closure $\overline{U}$. However, in the situation I am working, I only need this:

Conjecture. There is a finite set $A\subset\mathbb{C}$ so that $f$ admits a continuous extension to $\overline{U}\setminus A$.

Is this conjecture true? If not, are there some modest assumptions we could impose on $U$ to make it true?

Thank you!