Prove that there are an infinite number of discontinuities on this function.

Discontinuity needs to be proved at single points, not intervals. You've already started saying there are an infinite number of points $1/n$ (which is obvious).

Look closely at the negation of continuity at $x_0$: $$\lnot\left(\forall\epsilon>0,\;\exists\delta>0,\;\forall x,\;\left|x-x_0\right|<\delta\Rightarrow\left|f(x)-f(x_0)\right|<\epsilon\right)$$ All the quantifiers reverse - "not for all" means "there is a counterexample", "not exists" means "every possibility doesn't work": $$\exists\epsilon>0,\;\forall\delta>0,\;\exists x,\;\left|x-x_0\right|<\delta\;\wedge\;\left|f(x)-f(x_0)\right|\not<\epsilon$$

This tells you what you actually need to do to prove discontinuity: instead of being told $\epsilon$ and having to pick $\delta$ to work, you get to pick $\epsilon$ first, then you have to make every $\delta$ not work.

In short, you need to find points arbitrarily close to $x_0$ that are more than $\epsilon>0$ away from $f(x_0)$ for some fixed $\epsilon$.

Taking $x_0=\frac{1}{n}$ and $\epsilon=\frac{1}{2}$ is clearly fine; you didn't word the last part correctly. What you were trying to say is that $$\forall k_n\in\left(\frac{1}{n+1},\frac{1}{n}\right),\;f(k_n)=0$$ which is fine. However, the actual statement you need is "give me $\delta$, I will find you $k_n$ such that it breaks the definition of continuity at $x_0$".

The correct statement (let $x_0=\frac{1}{n},\;\epsilon=\frac{1}{2}$): Let $\delta>0$ be given. Choose $$k_n=\max\left(x_0-\frac{\delta}{2},\frac{1}{n+\frac{1}{2}}\right)$$ such that $$\left|k_n-x_0\right|<\delta\text{ and }\frac{1}{n+1}<k_n<\frac{1}{n}\text{.}$$ The first half says this is a valid $x$ for testing the continuity condition. The second half says $f(k_n)=0$ and therefore $\left|f(k_n)-f(x_0)\right|>\epsilon$.

Let $x_n = \frac{1}{n}$. Fix $n$ and consider $y_m = \frac{1}{n} + \frac{1}{(n+1)^m}$ with $m \geq 2$. Notice that $y_m = \frac{(n+1)^m + n}{n(n+1)^m}$ is a reduced fraction: \begin{eqnarray} \gcd((n+1)^m+n, n(n+1)^m) &=& \gcd((n+1)^m+n,n) \, \gcd((n+1)^m+n,(n+1)^m) \\ &=& \gcd((n+1)^m,n)\,\gcd(n,(n+1)^m) \\&=& 1 \end{eqnarray} Hence it's not possible to write it as $\frac{1}{k}$ for any $k$. Thus $f(y_m) = 0$ for each $m$ and $f(y_m) \to 0$. On the other hand $y_m \to x_n$ and $f(x_n) = 1$. Hence, $\lim_{m \to \infty} f(y_m) \neq f(\lim_{m \to \infty} y_m)$ and $f$ is not continuous at $x_n$.