Let me add a little to Arturo's answer.
First: how do you approach to problem?
You may have already known that a continuous function on any compact space is necessarily uniformly continuous. You may also know that a closed interval is a compact space. (Or you may just know that a continuous function on a closed interval is necessarily uniformly continuous.)
Therefore, if your continuous function $f:(0,1)\to\mathbb{R}$ admits limits
$ f_0 = \lim_{x\to 0} f(x) \qquad f_1 = \lim_{x\to 1} f(x) $
then defining the function
$ \tilde{f} :[0,1] \to\mathbb{R} $
by $\tilde{f}(0) = f_0$, $\tilde{f}(1) = f_1$ and $\tilde{f}(x) = f(x)$ otherwise you get a continuous function on $[0,1]$, and so $\tilde{f}$ must be uniformly continuous. And so $f$, being the restriction of $\tilde{f}$, must also be uniformly continuous.
This tells us that if you want to look for possible problems, you must consider the case where one of the limits, say $\lim_{x\to 0}f(x)$, fail to exist.
How can the limit fail to exist? There are basically two ways: either the $\limsup$ or the $\liminf$ do not exist (in which case the function is unbounded, and which you ruled out by assumption), or the $\limsup$ and $\liminf$ both exist but do not equal one another. Using the definition of the $\limsup$ and $\liminf$ you then "immediately see" that the oscillation that Arturo mentioned can cause problems to uniform continuity.
Note that this also explains the remark someone else made to you about monotonic functions: if you assume that $f$ is monotonic near $0$ and $1$, then you can use the fact that for bounded monotonic functions, the $\lim$ must exist, to conclude that $f$ extends to a continuous function on $[0,1]$ (like above) and therefore must be uniformly continuous.