Let $E$ be a closed, bounded set and let $f(z)$ be a continuous complex function in $E$. Prove that $f(z)$ is bounded in $E$.
I began the argument the same way that the boundness theorem is addressed in Real Analysis (believing that the argument shouldn't change much).
Assume $f(z)$ is not bounded on $E$. Then $\forall n \in \mathbb{N}, \space\space\exists z_n \in E \space\space$ s.t. $\space |f(z_n)| > n$. Construct the sequence $(z_n)_{n=1}^\infty \subset E \space$ from these $z_n$.
Note that $(z_n)_{n=1}^\infty$ is bounded, as $E$ is bounded.
Then by Bolzano-Weierstrass, $(z_n)_{n=1}^\infty$ has a limit point $L$, and so there exists a subsequence $(z_{n_k})_{k=1}^\infty$ which converges to $L$. $\space$ Moreover, $\space L \in E \space$ since $E \space$ is a closed set.
This implies $\lim_{k \to \infty} f(z_{n_k}) = f(L)$, $\space$ and so $\lim_{k \to \infty} |f(z_{n_k})| = |f(L)| \space$ because $f$ is continuous on $E$, and $f(z)$ continuous implies $|f(z)|$ is continuous.
Now, in real analysis proofs, we say this is a contradiction, as $\lim_{k \to \infty} |f(z_{n_k})| = \infty \space$ by assumption. However, we have introduced the extended complex numbers in my complex analysis class, and I don't see why $|f(L)| \ne \infty$ would be necessary, which is the crux of the contradiction.
So my question is:
If working in the extended complex numbers, would this proof method fail as I believe it would, or am I missing something? If it does fail, does this also mean that the theorem would not be true, or just that this is an insufficient way to prove it?
Your proof is fine because the absolute value of a complex number is a non-negative real number.
So indeed $|f(z_{n_k})| \to +\infty$
Note that if $f(z_{n_k}) \to f(L)$ then $|f(z_{n_k})| \to |f(L)|$
thus $|f(L)|$ must be finite because of the sequential continuity of $f$.