Function with countably many points of discontinuity Aside from rigor, is this proof correct?
Claim. Let $f$ be a function defined on $[0, 1]$ such that $\lim\limits_{y\to a} f(y)$ exists for all $a \in [0, 1]$. Then for any $\epsilon > 0$ there are only finitely many points $a \in [0, 1]$ with $$|\lim\limits_{y\to a} f(y) - f(a)| > \epsilon.$$
Proof. Suppose that there are infinitely many such points $a.$ Then by the Bolzano-Weierstrass Theorem, these points have a limit $x \in [0, 1].$ Let $$L := \lim\limits_{y \to x} f(y) = \lim\limits_{a\to x} f(a).$$
The condition $$|\lim\limits_{y\to a} f(y) - f(a)| > \epsilon$$ means that for $y$ close to $a$, $f(y)$ is far from $f(a).$ Similarly $\lim\limits_{a \to x} f(a) = L$ means that for $a$ close to $x,$ $f(a)$ is close to $L.$ Together this means that for $y$ close to $x$, $f(y)$ is far from $L$, but this contradicts the fact that $L = \lim\limits_{y \to x} f(y),$ i.e. for $y$ close to $x$, $f(y)$ is close to $L.$
 A: This proof is missing more than just "rigor"--there is some real subtlety in making the estimates you make at the end precise.  In particular, it is not necessarily true that $f(y)$ is far from $L$ for all values of $y$ sufficiently close to $x$.  Fortunately, in order to contradict the fact that $L = \lim\limits_{y \to x} f(y)$, you only have to show that $f(y)$ is far from $L$ for some values of $y$ arbitrarily close to $x$.
Let's take a closer look at your claims.  For any fixed $a$, $|\lim\limits_{y\to a} f(y) - f(a)| > \epsilon$ implies that there exists $\delta_a>0$ (possibly depending on $a$) such that $|f(y)-f(a)|>2\epsilon/3$ when $0<|y-a|<\delta_a$.  Also, $\lim\limits_{a \to x} f(a) = L$ implies that there exists a $\delta>0$ such that $|f(a)-L|<\epsilon/3$ whenever $0<|a-x|<\delta$.  Combining these two facts, you can conclude that whenever $0<|a-x|<\delta$ (and $|\lim\limits_{y\to a} f(y) - f(a)| > \epsilon$) and $0<|y-a|<\delta_a$, $|f(y)-L|>\epsilon/3$.
However, it is not necessarily true that every $y$ which is sufficiently close to $x$ satisfies $0<|y-a|<\delta_a$ for some $a$ such that $|\lim\limits_{y\to a} f(y) - f(a)| > \epsilon$ and $0<|a-x|<\delta$.  So you can't claim that $f(y)$ is far from $L$ for all $y$ sufficiently close to $x$.  What you can say is that for any $\delta'>0$, there exists $y$ such that $0<|y-x|<\delta'$ and $|f(y)-L|>\epsilon/3$: just choose $a$ such that $0<|a-x|<\min(\delta'/2,\delta)$ and $y$ such that $0<|y-a|<\min(\delta'/2,\delta_a)$.  This is then enough to reach a contradiction.
