Does the supremum norm $\|p\|_{A}$ depend continuously on subsets $A\subset\mathbb{C}$ with respect to the Hausdorff distance?

Question

Consider the space $\mathcal{K}$ of all non-empty compact subsets of $\mathbb{C}$. One can show that the Hausdorff distance defined by $$h(X,Y)=\max\bigg\{\sup_{x\in X}\inf_{y\in Y}|x-y|,\sup_{y\in Y}\inf_{x\in X}|x-y|\bigg\}\qquad\text{for all}\qquad X,Y\in\mathcal{K}$$ defines a metric on $\mathcal{K}$. Now suppose that we have a sequence $(A_{n})$ in $\mathcal{K}$ that converges to $A$ in $\mathcal{K}$ with respect to $h$. Can we conclude for any polynomial $p$ that $$\|p\|_{A_{n}}\to\|p\|_{A},$$ where $\|p\|_{A}:=\sup_{z\in A}|p(z)|$ is the usual supremum norm?

So we have to prove that $$\big|\|p\|_{A_{n}}-\|p\|_{A}\big|\to0.$$ Intuively, this result should be true. Any proof suggestions (or counterexamples) would be greatly appreciated.

Here are some observations:

Observation 1: By the maximum modulus principle we have $\|p\|_{A}=\|p\|_{\partial A}$, where $\partial A$ denotes the boundary of $A$.

Observation 2: By the reverse triangle inequality we have $$\big|\|p\|_{A_{n}}-\|p\|_{A}\big|=\big|\|1_{A_{n}}p\|_{\mathbb{C}}-\|1_{A}p\|_{\mathbb{C}}\big|\leq\|(1_{A_{n}}-1_{A})p\|_{\mathbb{C}}\leq\|p\|_{A_{n}\cup A},$$ where $1_{A}$ is the indicator function of $A$.

Observation 3: As Ethan Dlugie mentioned, the result is probably true for any continuous function $\mathbb{C}\to\mathbb{C}$. Not just polynomials $p$. (Caution: Observation 1 is not necessarily true for this more general setting.)

Observation 4: According to an MSE post we have $$\big|\|p\|_{X}-\|p\|_{Y}\big|\leq\sup_{\substack{x\in X, \ y\in Y\\|x-y|\leq h(X,Y)}}|p(x)-p(y)|.$$

Answer 1

Fix a compact set $A\subset \Bbb C$ and $\varepsilon>0$. Let $C=\{x\in\Bbb C:\mathrm{dist}(x,A)\leq\varepsilon\}$. Then $C$ is compact and let $\delta$ be from the definition of uniform continuity of $p$ on $C$. Wlog we can assume that $\delta\leq\varepsilon$.

If $h(A,A')<\delta\leq\varepsilon$ then $A'\subset C$. Moreover, for any $x'\in A'$ there is $x\in A$ such that $|x-x'|<\delta$, so $$|p(x')|\leq |p(x)|+|p(x')-p(x)|<\|p\|_A+\varepsilon.$$ Therefore $\|p\|_{A'}\leq \|p\|_A+\varepsilon$.

Similarly, for any $x\in A$ there is $x'\in A'$ such that $|x-x'|<\delta$, so $\|p\|_{A}\leq \|p\|_{A'}+\varepsilon$.

This shows that $|\|p\|_{A'}-\|p\|_{A}|<\varepsilon$.