How to use Lebesgue Number for Complex Analysis Holder Inequality

Question

Problem: Let $f$ be analytic in an open set $U\subset \mathbb{C}$ and let $K\subset U$ be compact. Show that there exists a constant $C$ depending on $U$ and $K$ such that $$|f(z)| \leq C \left( \int_U |f|^2 \right)^{1/2}$$ for all $z\in K$.

Approach: It is clear that I need to invoke Cauchy's integral formula and Holder's inequality with $p=2=q$ to arrive at the result. I am trying to figure out how to appropriately choose the closed curve that we integrate into Cauchy's integral formula to get our desired result. I am guessing Lebesgue number lemma is a nice way to get rid of $|\xi - z|$ in the denominator.

What I want: $$\begin{equation*} \begin{split} |f(z)| & = \left| \frac{1}{2\pi i} \int_{|\xi - z| = \delta} \frac{f(\xi)}{\xi - z} d\xi \right|\\ & \leq \frac{1}{2\pi} \int_{|\xi - z| = \delta} \frac{|f(\xi)|}{|\xi - z|}d|\xi|\\ & \leq \frac{1}{2\pi} \left( \int_{|\xi - z| = \delta} |f(\xi)|^2 d|\xi| \right)^{1/2} \left( \int_{|\xi - z| = \delta} \frac{1}{|\xi - z|^2} d|\xi| \right)^{1/2}\\ & = \frac{1}{\delta \sqrt{2\pi}} \left( \int_{|\xi - z| = \delta} |f(\xi)|^2 d|\xi| \right)^{1/2}\\ & \leq \frac{1}{\delta \sqrt{2\pi}}\left( \int_U |f|^2 \right)^{1/2} \end{split} \end{equation*}$$

Issue: How do I appropriately choose the Lebesgue number $\delta$ that I used in the work above? Two ways I thought of:

1. Take an arbitrary small $\epsilon > 0$ and consider the open cover $\{ B_{\epsilon}(z) \}_{z \in K}$. The Lebesgue number lemma tells us that there exists $\delta >0$ (Lebesgue number) such that if $B_\delta (z) \subset K \Rightarrow B_\delta (z) \subset B_\epsilon(z_0)$ for some $z_0 \in K$.

2. Consider the open cover $\{ U \}$ of $K$. The Lebesgue number lemma tells us that there exists $\delta >0$ such that if $B_\delta (z) \subset K \Rightarrow B_\delta(z) \subset U$ (which is trivially true since $K\subset U$ so this case doesn't seem appropriate).

I am confused because the Lebesgue number lemma expects us to satisfy $B_\delta (z) \subset K$. But $K$ is compact so if we consider a point $z'\in K$ on the boundary, then we can't say that there is a $\delta >0$ such that $B_\delta (z') \subset K$. What am I misunderstanding here? Thank you.

EDIT: Would the solution to my issues above be to consider an open cover $\{ B_\epsilon (z) \}_{z\in K}$ where $B_\epsilon (z) := B_\epsilon (z) \cap K $ is the open ball of radius $\epsilon$ in K as opposed to in $U$? It seems like taking the subspace topology will solve all the issues and we can bring back to the ambient topology in $U$ at the end of our proof. But this also seems pretty sketchy if there is an isolated singleton point in $K$...