Let $f$ be continuous prove that if $f:[a,b] \rightarrow \mathbb{R}$ then $|f(x)| \leq M$ for $M >0$ Let $f$ be continuous prove that if $f:[a,b] \rightarrow \mathbb{R}$ then $|f(x)| \leq M$ for $M >0$. Hint: suppose there exists a sequence $x_n \in [a,b]$ such that $\lim_{n \rightarrow \infty} |f(x_n)| = +\infty$
Note: I am aware that you compact sets get mapped to compact sets, but I am explicitly not allowed to use that result. The only result with regards to continuity I can use is the notion of sequential continuity. 
Attempt: ( I feel that my reasoning in the solution is a little flawed, feedback would be appreciated)
Suppose there exists a sequence $x_n \in [a,b]$ such that $\lim_{n \rightarrow \infty} |f(x_n)| = +\infty$
Given the sequence $x_n \in [a,b]$ and [a,b] is a bounded set, by the Bolzano Weitstrauss Theorem there exists a convergent subsequence $x_{n_{k}}$ such that $x_{n_{k}} \rightarrow x$ for $x \in [a,b]$
By continuity and the continuous mapping theorem if $x_n \rightarrow x$ then $f(x_n) \rightarrow f(x)$. In our case this means $f(x_{n_{k}}) \rightarrow f(x)$
By assumption $f(x_n)$ diverges, which means every subsequence must diverge. But we have a convergent subsequence from this sequence. This is a contradiction.
 A: Your reasoning isn't flawed but it's arguably incomplete.  I'll pick things up from where you left off.
Assume that no such $M$ exists.  Then $\forall n \in \Bbb N ~\exists x_n \in [a, b] \text{ such that } |f(x_n)| \gt n.$  Then $\lim_{n \to \infty} |f(x_n)| = \infty$, which we have just shown is impossible.  Thus, $\exists M~\forall x \in [a, b]~|f(x)| \lt M$.
A: You can also do a construction proof for  $M$.
Let $x_0 \in [a,b] $ and $ (x_n) $ be a sequence in $[a,b]$ where $\lim x_n = x_0$. Since $f$ is continuous, then clearly $\lim f(x_n) = f(x_0)$. This implies the sequence $(f(x_n))$ is convergent and consequently bounded. This is true for all convergent sequences in $[a,b]$. Then by the continuity of $f$, the image of every convergent sequence in $[a,b]$ is also convergent and consequently bounded. Choose $M$ to be the supremum of the upper bounds of the absolute value of the images of all convergent sequences in $[a,b]$. There's likely a cleaner way to word that. Therefore $|f(x)| \leq M $ for any $x \in [a,b]$.
