If $A$ is Compact, all Continuous Functions on $A$ are Bounded There are a lot of question asking for the converse, but not this direction (that I could find).

Let $A$ be a compact set, and let $V = \{ f:A \rightarrow \mathbb{R} ; \; f \text{ is continuous}  \}$. 
Claim: $$V = \{ f:A \rightarrow \mathbb{R} ; \; f \text{ is continuous} \; \}$$
$$ = V_b = \{ 
f:A \rightarrow \mathbb{R} ; \; f \text{ is continuous and bounded} \;\}.$$ 

I know that $A$ being compact means that all sequences have convergent subsequences. I don't know how to get from convergent sequences to bounded functions.
I also know that $A$ is totally bounded and complete, but that seems even less helpful.
 A: To expound on Chris' answer:
We know via the Heine-Borel theorem that compact sets in euclidean space are bounded. Then, it suffices to show that the image of a continuous function from $A$ is compact.
We have to use the compactness of $A$ somehow. You seem to be using the fact that all sequences have convergent subsequences. You probabally could do it this way, but its much easier to use the defining feature of compactness: every open cover has a finite subcover. 
Let $f:A\to \mathbb{R}$ be continuous, and $\mathscr{U}$ an open cover of $Im(f)$. Then, for each element $U \in \mathscr{U}$, we know $f^{-1}(U)$ is open. Because $\mathscr{U}$ is a cover of $Im(f)$, we know the collection of sets $\{f^{-1}(U) : U \in \mathscr{U}  \}$ covers $A$. By compactness of $A$, there is some finite subcover $\mathscr{I}$. Because $\mathscr{I}$ covers $A$, the set $\mathscr{S} = \{f(S) : S \in \mathscr{I}\}$ covers $Im(f)$. Then, $\mathscr{S}$ is a finite subcover of $\mathscr{U}$, meaning $Im(f)$ is compact.
A: The continuous image of a compact set is compact.  But in $\Bbb R$ compact implies bounded.
