There exists polynomial $(P_{n})_{n\in\mathbb{N}}$ such that $P_n(0)=1$, and $\lim_{n\rightarrow \infty}P_{n}(z)=0$..

Show that there exists a sequence of polynomial $(P_{n})_{n\in\mathbb{N}}$ such that $P_n(0)=1$ for each $n$, and $\lim_{n\rightarrow \infty}P_{n}(z)=0$ for all $z\in \mathbb{C}\setminus \left\{0\right\}$.

Remark: There are proofs of this fact that use versions of the Runge's theorem, the problem is that those versions do not coincide with versions I was taught in my course. Runge's theorem on my course is:

Runge's Theorem for compacts: Let $U\subseteq \mathbb{C}$ be an open set, $K\subseteq U$ compact set, $f:U\rightarrow \mathbb{C}$ holomorphic function. Let $\varepsilon >0$ and let $A\subseteq \mathbb{C}\setminus K$ such that $A$ intersects any connected component of $\mathbb{C}\setminus K$. Then there is $g\in \mathcal{R}_{A}$ such that $\left\|f-g\right\|_{K}<\varepsilon$.

where $$\mathcal{R}_{A}:=\left\{f \: \mbox{rational function with each pole in } A\right\}$$ In my attempt I use the following lemma:

Lemma $\bigstar$: Let $K\subseteq \mathbb{C}$ be a compact set. Let $r>0$ such that $A:=\left\{z\in \mathbb{C}\: : \: |z|>r\right\}\subseteq \mathbb{C}\setminus K$ and $f\in \mathcal{R}_{A}$. Then $f$ can be approximated uniformly on $K$ by polynomials.

My attempt: For each $n\in \mathbb{N}$ consider $C_{n}:=\left\{z\in \mathbb{C}\: : \: \frac{1}{n}\leq \mbox{Re}(z) \leq n \: \mbox{ and }\: \frac{-1}{4n}\leq \mbox{Im}(z) \leq \frac{1}{4n}\right\}$. We define $$K_{n}:=\left\{z\in\mathbb{C}\: : \: z\in \overline{B_{n}(0) } \: \mbox{and} \: d(z,\mathbb{R}^{+})\geq \frac{1}{n}\right\}\cup \overline{B_{\frac{1}{4n}}(0)}\cup C_{n}.$$ Clearly $K_ {n}$ is compact. Now, we consider the open set $V_{n}:=\left\{z\in \mathbb{C}\: : \: \frac{1}{n}\leq \mbox{Re}(z) \leq n \: \mbox{ and }\: \frac{-1}{3n}<\mbox{Im}(z) < \frac{1}{3n}\right\}\cup B_{\frac{1}{3n}}(\frac{1}{n})\cup B_{\frac{1}{3n}}(n)$. We define $$U_{n}:=\left\{z\in\mathbb{C}\: : \: z\in B_{n+\frac{1}{3n}}(0) \: \mbox{and} \: d(z,\mathbb{R}^{+})> \frac{2}{3n}\right\}\cup B_{\frac{1}{3n}}(0)\cup V_{n}.$$ Note that $U_{n}$ is open and also $K_{n}\subseteq U_{n}$. All this is summarized in the following figure:

Now, we consider $A_{n}:=\left\{z\in \mathbb{C} \: : \: |z|>n+\frac{1}{3n}\right\}$ and $f_{n}:U_{n}\rightarrow \mathbb{C}$ defined by: $$f_{n}(z):=\left\{\begin{array}{rl}1 & \mbox{If }z\in B_{\frac{1}{3n}}(0) \\ 0 & \mbox{If }z\in U_{n}\setminus B_{\frac{1}{3n}}(0) \end{array}\right.$$ Note that $f_{n}$ is holomorphic in $U_{n}$. Therefore, by Runge's Theorem for compacts there is $g_{n}\in \mathcal{R}_{A_{n}}$ such that $\left\|f_{n}-g_{n}\right\|_{K_{n}}<\frac{1}{2n}$.

But, by Lemma $\bigstar$ for $r_{n}=n+\frac{1}{3n}$, $g_{n}$ can be approximated uniformly on $K_{n}$ by polynomials. So, there exist a polynomial $P_{n}$ such that $\left\|g_{n}-P_{n}\right\|_{K_{n}}<\frac{1}{2n}$.

Therefore, $$\left\|f_{n}-P_{n}\right\|_{K_{n}}<\frac{1}{n} \: \mbox{ for all } n=1,2,\ldots,n.$$ Therefore, $(P_{n})_{n\in\mathbb{N}}$ is a sequence of polynomial such that $P_n(0)=1$ for each $n$, and $\lim_{n\rightarrow \infty}P_{n}(z)=0$ for all $z\in \mathbb{C}\setminus \left\{0\right\}$.

Questions: I want to know if there is any mistake in my test. In addition, I want to know if I am correctly applying the theorems statements.

If my proof has mistakes I would like you to give me some hint to complete the proof correctly.