Suppose that $(z_{n})_{n=1}^{\infty}$ is a sequence in $\textbf{C}$ which converges to $z$. Show that $\overline{z}_{n}\to\overline{z}$ and $|z_{n}|\to |z|$ as $n\to\infty$.
MY ATTEMPT
Let $\varepsilon > 0$. Then there is a natural number $n_{0}\geq 1$ s.t. $|z_{n} - z| < \varepsilon$ whenever $n\geq n_{0}$.
Moreover, one has that $|z_{n} - z| = |\overline{z_{n} - z}| = |\overline{z}_{n} - \overline{z}| < \varepsilon$. Thus we conclude that $\overline{z}_{n}\to \overline{z}$ as $n\to \infty$.
Similarly, one has that $||z_{n}| - |z|| \leq |z_{n} - z| < \varepsilon$. Thus $|z_{n}|\to|z|$ as $n\to\infty$, and we are done.
Could someone please check if the wording of my proof is good enough? Any comments are appreciated.
Yes, this is fine. By the way, for the second part it suffices to note that $\lvert \:\cdot\:\rvert:\Bbb{C}\to \Bbb{R}$ is continuous, so that it commutes with limits and $\lim_{n\to\infty} \lvert z_n\rvert=\lvert \lim_{n\to\infty} z_n\rvert=\lvert z\rvert$.