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 $\mathbb{R}^2\setminus \{ 0\}$.
Extend $f$ to $\hat f:\operatorname{cl}V\to\Bbb R$ just as you’ve already done. If $V$ is clopen in $X$, let
$$f^*:X\to\Bbb R:x\mapsto\begin{cases}
\hat f(x),&\text{if }x\in V\\
0,&\text{otherwise}\;.
\end{cases}$$
Otherwise, apply Uryson’s lemma to the normal space $\operatorname{cl}V$ to get a continuous $g:\operatorname{cl}V\to[0,1]$ such that $g(x)=1$ for all $x\in K$, and $g(x)=0$ for all $x\in(\operatorname{cl}V)\setminus V$. Then define
$$f^*:X\to\Bbb R:x\mapsto\begin{cases}
g(x)\hat f(x),&\text{if }x\in\operatorname{cl}V\\
0,&\text{otherwise}\;.
\end{cases}$$
Best Answer
For metric spaces you can do another approach as well (yours is very specific for the reals), following Engelking exercise 4.1F, but the result is due to Hausdorff.
Let $(X,d)$ be metric and suppose $A$ is closed in $X$, and $f: A \rightarrow [0,1]$ is continuous. Then define
$$F(x) = \begin{cases} f(x) & \text{if } x \in A \\ \inf{\left\{f(a) + \frac{d(x,a)}{d(x,A)} - 1\,:\,a \in A\right\}} & \text{if } x \in X \smallsetminus A\end{cases}$$
and show that $F$ is continuous.
(For unbounded functions use the $\arctan(f)$ to make it bounded first.)