Proving Existence of Discontinuity I need to prove that $f:[0, 1] \to \Bbb R$ given by $f(x) = \begin{cases}
1,  & \text{if $x=\frac{1}{n}$ for any positive integer $n$} \\
0, & \text{otherwise}
\end{cases}$ has an infinite number of discontinuities.
I've identified that the discontinuities exist at $x=\frac{1}{n}$ for positive integers $n \ge 2$.
My first attempt included trying to use the epsilon-delta definition, however, I've figured it'd be easier to use the limit definition (if $f$ is continuous at $x_0$, then $\lim\limits_{x \to x_0} f(x) = f(x_0)$). 
I'm just not sure how to satisfy the 'infinite' aspect of the question. I figure I need to prove the existence of one of the discontinuities and show how there exist many more of its kind?
Any hints into the right direction from here would be much appreciated.

Edit: It has been recommended I prove this using the epsilon-delta definition of continuity and do a proof by contradiction.
Thus, I claim $f$ is continuous, and for any $\epsilon >0$, there is a $\delta >0$ such that $$|x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\epsilon $$
Since $f$ is not actually continuous, I will contradict myself and show that the second part of the above implication fails. So, $|f(x)-f(x_0)|\ge \epsilon$.
I am not sure which such $\delta>0$ will achieve this? Any suggestions on how I can work that out?
 A: Let $n >4$. Then $\frac 1 n +\frac 1 {\sqrt 2},\frac 1 n +\frac  1 {2\sqrt 2},\frac 1 n +\frac  1 {3\sqrt 2},...$ is  a sequence in $[0,1]$ converging to $\frac 1  n$. What happens to the values of $f$ at these points? [Note that $\frac 1 n +\frac 1 {m\sqrt 2}$ can never be of the  form $\frac 1 k$ for any integer $k$ in view of irrationality of $\sqrt 2$]. 
A: Let $a=1/n$ where $n$ is some specific positive integer. Then $f(a) =1$ and further every neighborhood of $a$ contains points $x$ not of the form $1/n$ (prove this) so that $f(x) =0$ and thus $|f(x) - f(a) |=1$ for some points in every neighborhood of $a$. Thus the condition of continuity can not be satisfied for $\epsilon<1$ and $f$ is discontinuous at $a$.

You should also understand that not every proof in analysis is difficult. A major part of the difficulty is the unnecessary use of symbolism / formalism in proofs (some instructors insist on these and are part of the problem). Move past that and write your arguments in natural language with bare mimimum of math symbols and you will see that the difficulty of proofs is mostly apparent / superficial.
Also if a proof is difficult there are chances that the textbooks will offer it or at least give hints. 
