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?

Or try to go directly.  
Assume the set $D$ of discontinuities is uncountable, select a condensation point of $D$ (a point where every neighborhood is $D$ is uncountable), then show that a one sided limits fails to exist there.  I suggest the Cauchy criterion for nonexistence of a limit...
** more complete version **
If $f$ is discontinuous at a point $c$, then there is $\epsilon > 0$ so that for every $\delta>0$ there exists $x$ with $|x-c|<\delta$ and $|f(x)-f(c)|>\epsilon$.  Let $D_\epsilon$ be the set of points satisfying this for a given $\epsilon$.  The set of all discontinuities is
$$
D = \bigcup_{\epsilon>0} D_\epsilon = \bigcup_{n=1}^\infty D_{1/n}
$$
Suppose $D$ is uncountable.  Then $D_{1/n}$ is uncountable for some $n$.  Fix such an $n$.  An uncountable set in $\mathbb R$ has an accuulation point, say $c$ is an accumulation point of $D_{1/n}$.  Either every right-neighborhood of $c$ meets $D_{1/n}$ or every left-neighborhood of $c$ meets $D_{1/n}$.  
Suppose every left-neighborhood of $c$ meets $D_{1/n}$; we will show that the left limit of $f$ does not exist at $c$.  (The case for right-neighborhoods is similar, then the right limit does not exist.)  Indeed, for any $\delta>0$, the interval $(c-\delta, c)$ meets $D_{1/n}$.  Let $x \in (c-\delta,c)\cap D_{1/n}$.  There is $\delta'>0$ so that $(x-\delta',x+\delta') \subset (c-\delta,c)$.  By the definition of $D_{1/n}$, there is $y$ so that $|x-y|<\delta'$ and $|f(x)-f(y)|>1/n$.  Summary:
For every $\delta>0$ there exists $x,y$ so that $c-\delta < x < c$, $c-\delta< y < c$, but $|f(x)-f(y)|>1/n$.
Thus the Cauchy criterion for existence of the left-hand limit
$$
\lim_{t \to c^-} f(t)
$$
fails.
