There is a simple theorem in analysis that says any convergent sequence is bounded. This seems almost a triviality since if a sequence $(x_n)$ converges to a point $x$,
By definition of convergence every neighborhood of $x$ contains almost all (i.e., all but a finite number) terms of the sequence. Thus, for any $\varepsilon_n > 0$ there exists $N \in \mathbb{N}$ such that $n > N \implies x_n \in \mathbb{B}_{\varepsilon_n}(x)$
For the remaining $n - 1$ terms of the sequence, choose finite neighborhoods $\varepsilon_1, \dots \varepsilon_{n-1}$ about $x_1, \dots x_n$. Consequently, $ (x_n) \in \bigcup_{k=1}^n \mathbb{B}_{\varepsilon_k}(x_k) $
Since each $\varepsilon$-ball is a bounded set the union of the $n$ balls is a bounded set and since the image of the sequence is contained within this union it, too, is bounded.
This seems to constitute a proof. However, the proofs I've seen of this usually end up invoking the triangle inequality, choosing mysterious values of $\varepsilon$, etc. So, my question is, is the reasoning I outlined above a sound argument or is there something I'm overlooking?