Question
Is the proof below for the boundedness of convergent sequences valid, if a little convoluted? Statement (2) seems to be standing on some shaky ground.
Proof candidate
Suppose that some sequence $\{z_n\}$ converges to $z$, which means that $|z_n - z| < \varepsilon$ for all $n > N(\varepsilon)$, where $\varepsilon$ is an arbitrary positive real number. Let's call this statement (1).
Suppose, also, that the sequence $\{z_n\}$ is unbounded, which means that $|z_n| > M$ for all $n > N(M)$ for any given $M>0$. This is statement (2).
From statement (1) and the triangle inequality, we have $$ \varepsilon > |z_n - z| \geq |z_n| - |z|\\ \varepsilon > |z_n| - |z| $$ and from statement (2) we have $|z_n| - |z| > M - |z|$.
Combining these, we can write $$ M-|z| < \varepsilon $$ for all $n>\max\{N(\varepsilon),N(M)\}$. Choosing $M = K|z|$, where $K > 1$, we have $$ |z|(K - 1) < \varepsilon $$ which can't possibly be satisfied for any $\varepsilon$, therefore $\{z_n\}$ must be bounded for it to converge to $z$
As pointed out in the comments, your proof has a flaw in the statement (2).
Here is a direct proof :
Let $N$ an integer, since $z_n$ is convergent, there is $\varepsilon>0$ such that $|z_n-z|< \varepsilon$ for all $n>N$.
Therefore, $|z_n|<|z|+\varepsilon$ for all $n>N$ and let $M = \max_{n \in \{1,N\}} |z_n|$.
Finally, for all $n$, $|z_n| < \max(M,|z|+\varepsilon)$, which is finite. Hence, $(z_n)$ is bounded.