For all continuous functions it is true that $f:[a,b] \mapsto \mathbb{R}$ (with $a < b$) is bounded from above.
The question is to use the opposite position of that statement as well as to use the Weierstrass Interval technique for a suitable sequence.
Could somebody please provide a nice explanation/proof as well as the general idea?
Let $f : A → R$ be continuous on A. If K ⊆ A is compact, then f(K) is compact as well.
Proof: Let $(y_n) $be an arbitrary sequence contained in the range set $f(K)$. To prove this result, we must find a subsequence ($y_{n_k}$ ), which converges to a limit also in $f(K)$. The strategy is to take advantage of the assumption that the domain set K is compact by translating the sequence ($y_n$)—which is in the range of f—back to a sequence in the domain K. To assert that ($y_n$) ⊆ $f(K)$ means that, for each $n ∈ N$, we can find (at least one) $x_n ∈ K $with $f(x_n) = y_n$. This yields a sequence ($x_n$) ⊆ $K$. Because $K$ is compact, there exists a convergent subsequence ($x_{n_k}$ ) whose $limit x = lim x_{n_k}$ is also in K. Finally, we make use of the fact that f is assumed to be continuous on $A$ and so is continuous at x in particular. Given that ($x_{n_k}$ ) $→ x$, we conclude that ($y_{n_k}$ ) $→ f(x)$. Because $x ∈ K$, we have that $f(x) ∈ f(K)$, and hence $f(K)$ is compact.
Heisenberg's proof works in nice generality, but we can prove things more directly here.
Suppose that $f$ is unbounded. Then there is some sequence $(x_n)_{n=1}^\infty$ contained in the interval $[a,b]$ such that $\lim_{n\to\infty} f(x_n) = \infty$.
Now, consider the sequence $(x_n)$. If it is convergent, then it converges to some element within the interval $[a,b]$. That is, $\lim_{n\to\infty}x_n = \alpha$ for some $a \leq \alpha \leq b$. However, we then have that
$$ \infty = \lim_{n\to\infty} f(x_n) = f\Big(\lim_{n\to\infty}x_n\Big) = f(\alpha) $$
since $f$ is continuous. However, this would imply that $f$ is not defined on the whole interval $[a,b]$! Hence this cannot happen.
Now, $(x_n)$ may not be convergent itself. However, it will (since it is a bounded sequence) have a convergent subsequence, and so you can apply this same argument to that subsequence to obtain the desired conclusion.
