This is quite a difficult problem. I think the easiest way to attack it is to use a proof by contradiction.

So suppose that the result is false, and that for each real number t, the set is bounded. For each natural number m, let . Then is closed (not too hard to prove), and .

Now comes the hard part. Suppose that contains an interval , where . Then for all and for all integers n, . But as n increases, the intervals start to overlap. In fact, this will happen as soon as n is sufficiently large that , or . But that means that every sufficiently large real number x is in one of the intervals and so . It is easy to conclude from that that f is bounded. Since we are told that f is unbounded, it follows that cannot contain an interval. And since is closed, that means that is nowhere dense.

Thus is a countable collection of nowhere dense sets whose union is the whole of . But that is impossible by the Baire category theorem. That contradiction completes the proof.

Here is a very similar problem. Let f be a continuous function and suppose that for every t>0 the sequence f(t), f(2t), f(3t), ... tends to 0. Prove that f(x) tends to 0 as x→∞. The proof of that result also relies on the trick of using overlapping intervals and the Baire category theorem. There is a long and careful discussion of that problem in this entry in Tim Gowers's blog.