 Solution

By the Heine-Borel theorem, a noncompact subset $X\subseteq \mathbb{R}$ is either unbounded or not closed. If $X$ is unbounded, the restriction $f(x)=x$ is a continuous function with unbounded range.

Otherwise $X$ is bounded and not closed. Choose $a\in \overline{X}\setminus X$, and defineIts denominator is nonzero on $X$, so it is continuous. Since $a$ is in the closure, for every $n$ there is $x_n\in X$ with $0<|x_n-a|<1/n$. Hence $f(x_n)>n$, proving that this continuous function is unbounded. These two cases establish unbounded continuous witnesses to noncompactness of a real subspace.