$X$ compact, limit of $f$ exists for all $p\in X$ then $f$ uniformly continuous on $X$ minus a countable subset of $X$. 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:
Since $\lim_{q\to p}f(q)=L_p<\infty$, choose a $\delta_p$ such that $\left|f(q)-L_p\right|<\frac{\epsilon}{2}$ if $|q-p|<\delta_p$. 
Let $$E_p=\left\{q\in X\mid \left|q-p\right|=\frac{1}{2}\delta_p,\;p\in X\right\}$$
then the collection of all $E_p$ 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 $p\in E_{p_i}$ ($1\leq i\leq k$) hence $$|p-p_i|<\delta_{p_i}$$ and $$|q-p_i|\leq|p-p_i|+|p-q|<\delta+\frac{1}{2}\delta_{p_i}\leq\delta_{p_i}$$ 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? 
 A: I approached the problem with sequences; perhaps it will help in your approach.
For $x\in X,$ let $L(x)$ denote the limit of $f$ at $x,$ the limit taken in $X\setminus \{x\}.$ Set $E=\{x\in X: |f(x)-L(x)| >0\}.$ Note that $f$ is continuous on $X\setminus E.$ Note also that $E = \cup_n E_n,$ where $E_n=\{|f(x)-L(x)| >1/n\}.$ 
Claim: Each $E_n$ is finite. Proof: Suppose some $E_n$ is infinite. Then $E_n$ contains a sequence of distinct points $(x_m).$ Because $X$ is compact, some subsequence $x_{m_k}$ converges to some $x\in X.$ WLOG, $x_{m_k}\ne x$ for all $k.$ For each $k$ there is $y_k\in X,y_k\ne x,$ such that $|y_k-x_{m_k}| < 1/k$ and $|f(y_k) -L(x_{m_k})| < 1/2n.$ It follows that
$$\tag 1  |f(x_{m_k}) - f(y_k)| \ge |f(x_{m_k}) - L(x_{m_k})| - |L(x_{m_k})-y_k| \ge 1/n-1/2n = 1/2n$$
for all $k.$ That is impossible, because both sequences $x_{m_k},y_k$ approach $x$ in $X\setminus \{x\},$ and since $f$ has a limit at $x,$ the expression in $(1)$ must converge to $0.$ This proves the claim.
Since each $E_n$ is finite, $E = \cup_n E_n$ is countable. So we'll be done if we show $f$ is uniformly continuous on $X\setminus E.$ Let's prove that: If not, there is $\epsilon>0$ and sequences $x_m,y_m\in X\setminus E$ such that
$$\tag 2|x_m-y_m| <1/m \text { and } |f(x_m)-f(y_m)|>\epsilon$$
for all $m.$ Now some subsequence $x_{m_k}$ converges to some $x\in X.$ The subsequence $y_{m_k}$ must therefore also converge to $x.$ Could $x$ lie in $X\setminus E?$ No, because $f$ is continuous on $X\setminus E,$ and $(2)$ would show we have a contradiction. Could $x$ lie in $E?$ No, for same reason as in $(1)$: both sequences $x_{m},y_m$ approach $x$ in $X\setminus \{x\},$ and since $f$ has a limit at $x,$ we have $|f(x_m)-f(y_m)|\to 0.$ We violate $(2)$ again, and we're done.
