Suppose $A\subset \mathbb{R}$ is domain of function $f(x)$ on which it is uniformly continuous and $f(x)$ has property that for each $x_0 \in \overline A -A$ , $$\lim_{x \to x_0^-} f(x)= \lim_{x \to x_0^+}f(x).$$
Lets extend the domain of $f(x)$ to $\overline A$, by writing $f(x_0)=\lim_{x \to x_0}f(x)$ for each $x_0 \in \overline A -A$ , so that $f(x)$ is continuous on $\overline A$.
Will $f(x)$ be uniformly continuous on $\overline A$ ?
If cardianility of $\overline A -A$ is finite, then it is obviously true, so my question is about infinite case.
Yes, it is true. Clearly, $f$ continuous and $A$ is dense in $\overline A$. This, together with the fact that $f|_A$ is uniformly continuous, is enough to prove that $f$ is uniformly continuous.
Take $\varepsilon>0$. Since $f$ is uniformly continuous on $A$, there is a $\delta>0$ such that$$(\forall x,y\in A):|x-y|<\delta\implies\bigl|f(x)-f(y)\bigr|< \frac{\varepsilon}2.$$Now let $x,y\in\overline A$ be arbitrary with $|x-y|<\delta$. There are sequences $(x_n)_{n\in\mathbb N}$ and $(y_n)_{n\in\mathbb N}$ of elements of $A$ such that $\lim_{n\to\infty}x_n=x$ and that $\lim_{n\to\infty}y_n=y$. Since $|x-y|<\delta$, we can assume, without loss of generality, that $(\forall n\in\mathbb{N}):|x_n-y_n|<\delta$, which implies that $(\forall n\in\mathbb{N}):\bigl|f(x_n)-f(y_n)\bigr|<\frac\varepsilon2$. So,$$\bigl|f(x)-f(y)\bigr|=\lim_{n\to\infty}\bigl|f(x_n)-f(y_n)\bigr|\leqslant\frac\varepsilon2<\varepsilon.$$