There are two parts to this statement. The first half does not require completeness.
Claim If $f:(a,b)\to\mathbb{R}$ is uniformly continuous, $\limsup_{x\to a^+} f \leq \liminf_{x\to a^+} f < \infty$.
Sketch of proof First we show that $\limsup f$ must be bounded. Suppose it were not. Let $\delta$ be some small number such that $\delta < b-a$. Consider $x_0 = a + \delta$, and $y_0 = f(x_0)$. Construct a sequence $(x_i,y_i)$ in the following manner. Pick $x_i < a + \frac12 (x_{i-1} - a)$ such that $y_i = f(x_i) > 2 y_{i-1}$. (Notice that if this sequence terminates after finitely many terms and cannot be continued, you have will have that $f|_{(a,\frac12(x_N + a))}$ is bounded.) However, we have that $|x_{i+1} - x_{i}| < \frac{\delta}{2^i}$ while $|y_{i+1} - y_{i}| > 2^i y_0$. So it is easy to see that by taking $i\nearrow \infty$ you get a contradiction to uniform continuity.
Similarly $f$ must be bounded below. Now suppose the $\limsup f > \liminf f$ are not equal. Then by a similar argument, you can construct an alternating sequence x_1, x'_1, x_2, x'_2 such that 2 |x'_i - a| < |x_i - a| < \frac12 |x'_{i-1} - a| and $f(x_i) \to \limsup f$ while f(x'_i) \to\liminf f. Then you have that for $0 < \epsilon < \limsup f -\liminf f$, for any $\delta$, there exists some $x_i$ such with a point x'_i within $\delta$ of it taking value more than $\epsilon$ away under $f$, contradicting uniform continuity. Q.E.D.
Like Nate said in his comments, now you are required to use the completeness of $\mathbb{R}$. By using the completeness you can say that since $\limsup f \leq \liminf f$, the limit $\lim_{x\to a^+}f$ exists, so setting $f(a) = \lim_{x\to a^+} f$ you get a continuous function.
For illustration, imagine instead of you have the function $f:(0,1)\to \mathbb{R}^2\setminus \{ 0\}$ given by
$ f(s) = (s \cos s, s\sin s) $
This function is uniformly continuous on $(0,1)$, but does not have continuous extension to $[0,1)$ with codomain $\mathbb{R}^2\setminus \{ 0\}$.