Is the set of discontinuity of $f$ countable? Suppose $f:[0,1]\rightarrow\mathbb{R}$ is a bounded function satisfying: for each $c\in [0,1]$ there exist the limits $\lim_{x\rightarrow c^+}f(x)$ and $\lim_{x\rightarrow c^-}f(x)$. Is true that the set of discontinuity of $f$ is countable?

Hopefully this more tedious proof is more illustrative...
In light of the assumptions on $f$, there are two ways that $f$ can fail to be continuous: (1) the left and right hand limits differ, or (2) the limits are equal, but the function value differs from the limit (thanks to @GEdgar for pointing this out).
Let $\epsilon>0$ and $\Delta_\epsilon = \{ x |\, |\lim_{y \downarrow x}f(y) - \lim_{y \uparrow x}f(y)| \geq \epsilon \}$. If $\Delta_\epsilon$ is not finite, then since $[0,1]$ is compact, there exists an accumulation point $\hat{x} \in [0,1]$. Let $c_+ = \lim_{y \downarrow \hat{x}}f(y), c_- = \lim_{y \uparrow \hat{x}}f(y)$. (Note that it is possible that $c_- = c_+$.) By assumption, there exists some $\delta>0$ such that for $y \in (\hat{x}-\delta,\hat{x})$, $|f(y)-c_-| < \frac{\epsilon}{4}$ and for $y \in (\hat{x}, \hat{x}+\delta)$, $|f(y)-c_+| < \frac{\epsilon}{4}$. However, this implies that $\hat{x}$ is an isolated point of $\Delta_\epsilon$, which is a contradiction. Hence $\Delta_\epsilon$ is finite.
If we let $\Delta = \cup_n \Delta_{\frac{1}{n}}$, we see that $\Delta $ is at most countable.
@GEdgar has pointed out an omission in my proof: It is possible that the two limits coincide at a point, but the function is still discontinuous at that point, ie, $\Delta$ is not the entire set of discontinuities.
Let $\Gamma_\epsilon = \{ x |\, \lim_{y \downarrow x}f(y) = \lim_{y \uparrow x}f(y), \ |\lim_{y \downarrow x}f(y)-f(x) | \geq \epsilon \}$. Suppose, as above, that $\Gamma_\epsilon$ is not finite, and let $\hat{x}$ be an accumulation point. Let $c_+, c_-$ be the limits as above. Again, there exists a $\delta>0$ such that if $y \in (\hat{x}-\delta,\hat{x})$, $|f(y)-c_-| < \frac{\epsilon}{4}$, and similarly, if $y \in (\hat{x}, \hat{x}+\delta)$, $|f(y)-c_+| < \frac{\epsilon}{4}$. Consequently, $\hat{x}$ is isolated, hence a contradiction, and $\Gamma_\epsilon$ is finite.
If we let $\Gamma= \cup_n \Gamma_{\frac{1}{n}}$, we see that $\Gamma$ is at most countable.
Since the set of discontinuities is $\Delta \cup \Gamma$, we see that the set of discontinuities is at most countable.
