My course notes (Mathematics BSc, second year module in real analysis, unpublished) have the following proof of the first part of the Boundedness Theorem.
The Boundedness Theorem: If $f:[a,b]\rightarrow\mathbb{R}$ is continuous, then it is bounded on $[a,b]$ and it attains its bounds there.
Proof: We first show that $f$ is bounded. To do this, we'll assume that it isn't, and seek a contradiction. So assume $f$ is not bounded. Let $(x_n)$ be a sequence in $[a,b]$ such that $|f(x_n)|>n$ for each $n\in\mathbb{N}$. Such a sequence certainly exists; for example, to construct such a sequence, we could define $x_n=\text{inf}\{x\in[a,b] \; | \; |f(x)|>n\}$.
Now since $(x_n)$ is a bounded sequence (as $a \leq x_n \leq b$ for all $n\in\mathbb{N}$), it has a convergent subsequence $(x_{n_k})$ by the Bolzano-Weierstrass Theorem. Let $x=\lim_{k\rightarrow\infty}f(x_{n_k})$. Then the sequence $(f(x_{n_k}))$ is bounded as it's convergent, and this gives our required contradiction.
But where's the contradiction? We assumed $f$ was unbounded and showed that the sequence $(f(x_n))$ has a bounded subsequence. How does that tell us it has no unbounded subsequence?
You find a contradiction because you are assuming that $|f(x_n)|$ diverges to infinity; but then, you extract a convergent subsequence of $(x_n)$, namely $(x_{n_k})$, converging to some $c$ in $[a,b]$. Because $f$ is continuous, $|f(x_{n_k})|$ converges to $|f(c)|$, which is a totally well defined number in $\mathbb{R}$ (because $f(c)\in\mathbb{R}$), i.e. does not explode to infinity. To conclude, remember that if the limit of a sequence exists, then the limit of a subsequence is the same as the limit of the sequence. Therefore, in the end you say that $|f(x_n)|$ converges to $|f(c)|$, which is the contradiction you were looking for