Since the inverse image under $f$ of an open interval is an open interval, and inverse images preserve unions, the inverse image under $f$ of an open set is open. When I say that inverse images preserve unions, I mean that $$f^{-1}\left[\bigcup\mathscr{U}\right]=\bigcup_{U\in\mathscr{U}}f^{-1}[U]\;.$$ Thus, if each $U\in\mathscr{U}$ is an open interval, then each $f^{-1}[U]$ is an open interval, and $f^{-1}\left[\bigcup\mathscr{U}\right]$ is a union of open intervals and therefore an open set.
You can also very easily use the $\epsilon$-$\delta$ definition of continuity: for each $\epsilon>0$ just take $\delta=\epsilon$, and observe that if $|x-x_0|<\epsilon$, then $|f(x)-f(x_0)|<\epsilon$. (This actually shows that $f$ is not just continuous, but uniformly continuous.)