For simplicity, let's take our compact set $X$ to be a subset of $\mathbb{R}$. Define $f:X\to\mathbb{R}$ to have a limit at every point in $X$. Then $f$ is uniformly continuous on a set $X$ minus a countable subset of $X$.
Here is what I have so far:
Fix $\epsilon>0$. Since for every $p\in X$, $\lim_{q\to p}f(q)=L_p<\infty$, we can choose a a $\delta_p$ such that $\left|f(q)-L_p\right|<\frac{\epsilon}{2}$ for every $q\in X$ such that $|q-p|<\delta_p$.Let $$E_p=\left\{q\in X: \left|q-p\right|\leq \frac{1}{2}\delta_p\right\}$$then the collection $\{E_p\}_{p\in X}$ is an open cover of $X$ and since $X$ is compact we have a finite set of point $\{p_1,\ldots,p_k\}$ in $X$ such that $$X\subset E_{p_1}\cup\ldots\cup E_{p_k}.$$If we let $$\delta=\frac{1}{2}\min\{\delta_{p_1},\ldots,\delta_{p_k}\}$$and choose $p,q\in X$ such that $|p-q|<\delta$ then $p\in E_{p_i}$ for some $1\leq i\leq k$.Hence $$|p-p_i|<\delta_{p_i}$$ and therefore we also have $$|q-p_i|\leq|p-p_i|+|p-q|<\delta+\frac{1}{2}\delta_{p_i}\leq\delta_{p_i}$$ which tells us that $$|f(p)-f(q)|\leq|f(p)-L_{p_i}|+|f(q)-L_{p_i}|<\epsilon.$$
This tells that every $\epsilon>0$ generates finitely many points $p_1,\ldots,p_k$ and a cover $\{E_{p_1},\ldots,E_{p_k}\}$ of $X$ and a $\delta>0$ such that $|f(q)-f(p)|<\epsilon$ if $|p-q|<\delta$.Now someone hinted that I should construct a sequence $\{\epsilon_n\}$ of real numbers numbers that goes to $0$ as $n\to\infty$. This will give us a countable union of finite sets $\{p_1,\ldots,p_k\}$ for every $\epsilon_n$ at which the function is not necessarily continuous.
Can I get any help with completing the proof?
Ah I see the problem is that the inequality$$|f(p)-L_{p_i}|+|f(q)-L_{p_i}|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$is no loner guaranteed if one of $p$ or $q$ is equal to $p_i$. This is because if it happens that let's say $p=p_i$ and $f(p_i)\neq L_{p_i}$ (which is possible since $f$ can be discontinuous as $p_i$), then $|f(p)-L_{p_i}|<\epsilon/2$ is no longer true. So in fact, what we can only guarantee is that for all $p,q\in X\setminus \{p_1,\ldots,p_k\}$ such that $d(p,q)<\delta$, we have $|f(p)-f(q)|<\epsilon$. So now the proof becomes simple. WLOG, we need only to take $\epsilon_n$ a positive sequence decreasing to 0. For $\epsilon_1$ and following the reasoning above, we can find a finite subset $E_1\subset X$ points and a $\delta_1>0$ such that for all $x,y\in X\setminus E_1$ with $d(x,y)<\delta$ we have $|f(x)-f(y)|<\epsilon_1$. The going by induction, there we can show that for each $n$, there is a finite set $E_n$ and a $\delta_n>0$ such that for all $x,y\in X\setminus (E_1\cup\cdots\cup E_n)$ such that $d(x,y)< \delta_n$ we have $|f(x)-f(y)|<\epsilon_n$. Taking $E=\bigcup_n E_n$, the $f$ is clearly uniformly continuous on $X\setminus E$ and $E$ is at most countable being a countable union of finite sets.