The following problem is a problem of Riemann mapping theorem's section of Conway's book:
Let $G$ and $\Omega$ open sets in the plane. If $f: G \longrightarrow \Omega$ is a homeomorphism, $\{ z_n \} \subset G$ which $z_n \rightarrow z \in \partial G$ and $\lim f(z_n) = w$ exists, then $w \in \partial \Omega$.
I would like to know if my attempt is correct, because this section is about the Riemann mapping theorem, but I didn't use this theorem in my attempt and I don't see how use this theorem to prove the exercise.
$\textbf{My attempt:}$
Suppose by absurd that $w \notin \partial \Omega$, then $w \in \mathbb{C} \backslash \overline{\Omega}$ or $w \in \Omega$.
If $w \in \mathbb{C} \backslash \overline{\Omega}$, then $d(w, \overline{\Omega}) > 0$. Given $\varepsilon := \frac{1}{2}d(w,\overline{\Omega}) > 0$, we can find $N \in \mathbb{N}$ such that
$$n \geq N \Longrightarrow |f(z_n) - w| < \frac{1}{2}d(w,\overline{\Omega}),$$
which is an absurd since $f(z_n) \in \Omega \subset \overline{\Omega}$ for each $n \in \mathbb{N}$, which imply that $|f(z_n) - w| \geq d(w,\overline{\Omega})$.
If $w \in \Omega$, then exist $z^* \in G$ such that $f(z^*) = w$ and $z^* \neq z$.
Given $\delta > 0$ and $\varepsilon > 0$, exist $N \in \mathbb{N}$ such that
$$n \geq N \Longrightarrow |z_n - z| < \delta \Longrightarrow |f(z_n) - w| < \frac{\varepsilon}{2} \ (*)$$
Let be $\{ z_n' \} \subset G$ a sequence which $z_n' \rightarrow z^*$. By the continuity of $f$ at $z^*$, given $\varepsilon > 0$, exist $\delta' > 0$ such that
$$|a - z^*| < \delta' \Longrightarrow |f(a) - f(z^*)| = |f(a) - w| < \frac{\varepsilon}{2}$$
and, given $\delta'' > 0$, exist $N \in \mathbb{N}$ such that
$$n \geq N \Longrightarrow |z_n' - z^*| < \delta'' < \delta',$$
where we took $\delta'' < \delta'$ since $\delta'' > 0$ is arbitrary, then
$$n \geq N \Longrightarrow |z_n' - z^*| < \delta'' < \delta' \Longrightarrow |f(z_n') - w| < \frac{\varepsilon}{2} \ (**)$$
Since $\delta, \delta'' > 0$ are arbitrary, we can take them small enough such that $B(z, \delta) \cap B(z^*, \delta'') = \emptyset$, then $|z_n - z_n'| > |z - z^*| - \delta - \delta'' > 0$ $(***)$.
By $(*)$, $(**)$ and continuity of $f^{-1}$ at $w$, must exist an $\eta > 0$ such that
$$|f^{-1}(f(z_n)) - f^{-1}(f(z_n'))| < \eta \Longrightarrow |z_n - z_n'| < |z - z^*| - \delta - \delta'',$$
whenever $n \geq N$, which contradicts $(***)$, then $w \notin \Omega$. Since $w \notin \mathbb{C} \backslash \overline{\Omega}$ and $w \notin \Omega$, we must have $w \in \partial \Omega$. $\square$
Thanks in advance!
The idea is correct, as far as I can tell. But there are some imprecise formulations. For example:
What you probably mean is that given $\varepsilon > 0$, there is a $\delta > 0$ such that $$ |z_n - z| < \delta \Longrightarrow |f(z_n) - w| < \frac{\varepsilon}{2} \, , $$ and for this $\delta$ there is a $N \in \mathbb{N}$ such that $$ n \geq N \Longrightarrow |z_n - z| < \delta \, . $$
Having said that, the arguments can be simplified. In the first part you show that $w \in \overline\Omega$, and the reason is that $\overline\Omega$ is closed. So you can simply write $$ \forall n: f(z_n) \in \Omega \subset \overline\Omega \\ \implies w = \lim_{n \to \infty} f(z_n) \in \overline\Omega $$ because $\overline\Omega$ is closed.
In the second part you show that $w \in \Omega$ is not possible, and the reason is that $f^{-1}$ is continuous. You could write: If $w \in \Omega$ then $$ f(z_n) \to w \in \Omega \\ \implies z_n = f^{-1}(f(z_n)) \to f^{-1}(w) \in G $$ because $f^{-1}$ is continuous at $w$. This contradicts the assumption that $z_n \to z \in \partial G$.