Here's what I have to show:
Suppose that $f:[0,\infty )\to \mathbb{R}$ is continuous and that $(x_n)$ is a sequence in $[0,\infty )$ such that $\lim f(x_n)=+\infty $. Prove that $\lim x_n = +\infty$.
Here's my attempt:
Suppose that $(x_n)$ does not coverge to $+\infty $. Then there exists a subsequence $(x_{n_k})$ of $(x_n)$ which is (strictly) bounded above by some $M>0$. Thus, $x_{n_k} \in [0, M)$ for every $k\in \mathbb{N}$. By Bolzano Weierstrass Theorem, there exists a subsequence $(x_{n_{k^{\prime}}})$ of $(x_{n_k})$ converging to some $c\in [0,M]$. Now since $f$ is continuous on $[0,M]$, we have $\lim f(x_{n_{k^{\prime}}})=f(c)$. However, since $(f((x_{n_{k^{\prime}}}))$ is a subsequence of $(f(x_n))$, we also should have $\lim f(x_{n_{k^{\prime}}})=+\infty$. A contradiction!
Is my proof correct? My proof looks extremely clumsy. Perhaps I was wondering if we could prove it directly.
This is correct, but you don't need to use the Bolzano-Weierstrass theorem. Once you know a subsequence of $(x_n)$ is contained in $[0, M]$ for some $M$, you can just say that $f$ is bounded on $[0, M]$ by continuity.