The following is a partial reply to the intended question. It looks to me as if you unfortunately did not quote the question verbatim. And the header is rather misleading.
But it is clear that the original question is something like this. Suppose that $f$ is continuous. Show that if $|f(x)|\ge x$ for all $x$, then $\lim_{x\to\infty} f(x)=\infty \quad \text{or} \quad \lim_{x\to\infty} f(x)=-\infty$
You say that you can handle things if you can show that $f(x)$ is either (i) positive for all positive $x$ or (ii) negative for all positive $x$. It is indeed true that $f(x)$ is either positive for all positive $x$ or $f(x)$ is negative for all positive $x$. But to save a tiny amount of time, we will show that either $f(x)$ is positive for all $x \ge 1$ or $f(x)$ is negative for all $x \ge 1$. That is plenty enough to allow pushing the rest of the proof through.
There are two possibilities: (i) $f(1)$ is positive or (ii) $f(1)$ is negative. Note that we cannot have $f(1)=0$, for $|f(1)|\ge 1$ by the condition the problem put on $f$.
We deal with possibility (i). Dealing with possibility (ii) is essentially identical (or cheat: if $f(1)$ is negative, let $g(x)=-f(x)$.)
We will show that if $f(1)$ is positive, then $f(x)$ is positive for all $x >1$. Note that $f(x)$ cannot be $0$ at any $x >1$, for that would violate the condition $|f(x)|>$. We next show that $f(x)$ cannot be negative at any $x>1$.
For suppose to the contrary that $f(b)<0$, where $b>1$. Our function is positive at $1$, and supposedly negative at $b$. Since $f$ is continuous, by the Intermediate Value Theorem we can conclude that $f(c)=0$ for some $c$ between $1$ and $b$. This, as we observed already, is impossible because $|f(c)| \ge c$.
So in case (i), $f(x)>x$ for all $x \ge 1$, from which the statement about the limit as $x \to\infty$ follows easily.
I have written out case (i) in gruesome detail. Of course the argument can be streamlined, but be careful to preserve meaning when you do!