If $f$ is continuous on $[0,\infty)$ and not bounded above implies... $Conj:$ If $f$ is continuous on $[0,\infty)$ and not bounded above implies there exists $\{ x_n\}$ such that $x_n \rightarrow \infty$ and $f(x_n)\rightarrow \infty$.
I can see how to show $f(x_n) \rightarrow \infty$ but not $x_n \rightarrow \infty$:
$Proof:$
For any $M$ there exists $x_1$ such that $f(x_1)>M$ by hypothesis.  Now, choose $N = f(x_1)$. There exists $x_2$ such that $f(x_2)>f(x_1)>M$. Continue iteratively to construct $\{x_n\} = \{x_1, x_2, \dots\}$. Now for all $K>0$ exists $n^*$ such that $n>n^*$ implies $f(x_n)> K$. Thus $f(x_n) \rightarrow \infty$.
Now what? Thanks!

To show that $\{x_n\}$ tends to infinity we can use the following argument about compactness and continuous functions.
Since $[0,x]$ is a compact set $\forall x\in \mathbb{R}^+$, and $f$ is a continuous function from this set to $\mathbb{R}$, $f([0,x])$ must be compact and therefore closed and bounded. Then if it is the case that $f(x_n)\rightarrow \infty$, it follows $x_n\rightarrow \infty$.
