Finding a sequence of polynomials that converges uniformly to a holomorphic function on an open set The following is exercise 13.2 in Rudin's Real & Complex Analysis, which I'm self-studying.

Let $\Omega = \{z: |z| < 1 \text{ and } |2z - 1| > 1\}$, and suppose $f \in H(\Omega)$. Must there exist a sequence of polynomials which converges to $f$ uniformly in $\Omega$?

I have a solution, but I feel it's simple and the shape of $\Omega$ is suspicious so there might be a trick somewhere.
Assume such a sequence exists. Let $0 < \epsilon < 1$, $f(z) = 1/z$ and $P$ be a polynomial that satisfies:
$$|f(z) - P(z)| < \epsilon \ \ \forall z \in \Omega$$
Therefore
$$|P(z)| < |f(z)| + \epsilon \ \ \forall z \in \Omega$$
But near the boundary of the unit disc, $|f(z)| < 2$, so $|P(z)| < 3$ near the boundary. By the continuity of $P$, we have $P(z) \le 3$ on the boundary. By the maximum modulus principle, $P(z) < 3$ on the unit disc. But $|f(z)|$ gets arbitrarily large near $0$. Therefore $P$ cannot approximate $f$ on $\Omega$.
What gives me confidence in my argument is that it doesn't work on compact subsets of $\Omega$ (for which the existence of the polynomial sequence is guaranteed by Runge's theorem).
Is my counter-example correct?
 A: Let $\Omega$ as defined above and $\mathbb{D}$ be the unit disk.
Suppose that $p_n\rightarrow f$ uniformly on $\Omega$. 
$\textbf{Claim}$ 1:$p_n$ is uniformly cauchy on $\overline{\Omega}$ 
Proof: Let $w \in \overline{\Omega} $ and $\epsilon>0$ be given. 
By uniform convergence on $\Omega$,  $\exists$ $N>0$ s.t. $|p_n(z)-p_m(z)|\leq \epsilon/3$ for all $z\in \Omega$ and $n,m\geq N$.
By continuity of $p_n,p_m$ $\exists$ $z\in\Omega$ s.t. $|p_n(z)-p_n(w)|\leq \epsilon/3$ and $|p_m(z)-p_m(w)|\leq \epsilon/3$
Then for $w\in \Omega, n,m\geq N$ $|p_m(w)-p_n(w)|\leq |p_n(w)-p_n(z)|+|p_n(z)-p_m(z)|+|p_m(z)-p_m(w)| \leq \epsilon$. 
So, $p_n$ is a Cauchy sequence in $(C(\overline{\Omega}), sup)$
$\mathbf{Claim}$ 2: $p_n$ is a Cauchy sequence in $(C(\overline{\mathbb{D}}), sup)$
Proof: Since $\partial \mathbb{D}\subset \overline{\Omega}$. $p_n$ is uniformly cauchy on $\partial \mathbb{D}$. And by maximum modulus principle, we have that $p_n$ is uniformly Cauchy on $\mathbb{D}$. Thus $p_n$ is a Cauchy sequence in $(C(\overline{\mathbb{D}}), sup)$.
Finally, By completeness of $C(\overline{\mathbb{D}})$, $p_n\rightarrow f$ for some $f\in C(\overline{\mathbb{D}})$ and since $p_n$ is uniformly bounded in $\mathbb{D}$, by Hurwitz's theorem, $f$ is holomorphic in $\mathbb{D}$. i.e. $f \in H(\mathbb{D})\cap C(\overline{\mathbb{D}})$.
