There are 2 parts to this question:
Let $f$ be continuous on the closed interval $[a, b]$. Then $f$ is bounded on $[a, b]$.
Assume for contradiction that $f$ is unbounded. For any $M \in \mathbb{N} > 0$, there exists $c$ in $[a, b]$ such that $|f(c)| > M$.
Since $[a, b]$ is closed, there exists a sequence $\{x_k\}_{k=1}^{\infty}$ in $[a, b]$ that tends to $c$
Part A: (Have I really used closedness here? As any point in any set has a sequence which converges to it, for example take the constant sequence. I have adapted this from a statement that A set $A$ is closed if and only if $x \in A$ whenever $x_k \in A$ and $\lVert x_k – x \rVert \to 0$ as $k \to \infty$.)
By sequential continuity, $f(x_k)$ must tend to $f(c)$.
Now, aiming for a contradiction:
$ |f(x_k) – f(c)| \geq |f(x_k)| – |f(c)| > |f(x_k)| – M $
Then taking the limit as $k$ tends to infinity.
$ \epsilon > |f(c)| – M > 0$, which does not quite get the contradiction.
Part B: Is there a way to finish this proof this way, without using sequential compactness as I have seen in other proofs online?
Best Answer
Part A: No, you haven't really used closedness. You're correct that any point has the constant sequence converging to it. The key in the definition of closedness you've provided is that $x$ doesn't need to belong to $A$ in the first place: if you're given a point $x$, then having a sequence $x_n \rightarrow x$ with $\{x_n\}_n \subseteq A$ allows you to conclude that $x \in A$. In any case, closedness isn't enough for this proof - the interval needs to be closed and bounded.
Part B: I don't believe there's any way to salvage your proof. The problem is that you've just taken one arbitrary value of $M$. This doesn't use unboundedness - you're just taking a (possibly large) value of $f$. You really need to use the fact that you have a sequence of points $(c_n)_n$ such that $f(c_n) \rightarrow \infty$.