The following is a standard application of Baire Category Theorem:
Set of continuity points of point wise limit of continuous functions from a
Baire Space to a metric space is dense $G_\delta$ and hence can not be countable.
Another result is the following:
Any monotone function on a compact interval is a pointwise limit of continuous functions.
Such a function can have countably infinite set of discontinuities. For example in $[0,1]$ consider the distribution function of the measure that gives probability $1/2^n$ to $r_n$ where $(r_n)$ is any enumeration of rational numbers in $[0,1]$. The set of discontinuity points of this function is $\mathbb{Q}\cap[0,1]$.
This sometimes called the Popcorn Function, since if you look at a picture of the graph, it looks like kernels of popcorn popping. (Also called the Thomae's Function.)
To prove it is discontinuous at any rational point, you could argue in the following way: Since the irrationals that are in $(0,1)$ are dense in $(0,1)$, given any rational number $p/q$ in $(0,1)$, we know $f(p/q) = 1/q$. But let $a_{n}$ be a sequence of irrationals converging to $p/q$ (by the definition of density). Then $\lim \limits_{a_{n} \to (p/q)} f(x) = 0 \neq f(p/q)$. Thus, $f$ fails to be continuous at $x = p/q$.
Now, to prove it is continuous at every irrational point, I recommend you do it in the following way:
Prove that for each irrational number $x \in (0,1)$, given $N \in \Bbb N$, we can find $\delta_{N} > 0$ so that the rational numbers in $(x - \delta_{N}, x + \delta_{N})$ all have denominator larger than $N$.
Once you have the result from above, we can use the $\epsilon-\delta$ definition of continuity to prove $f$ is continuous at each irrational point.
So, first prove 1. (It's not too hard.) Once you've done that, you can accomplish 2. in the following way:
Let $\epsilon > 0$. Let $x \in (0,1)$ be irrational. Choose $N$ so that $\frac{1}{n} \leq \epsilon$ for every $n \geq N$ (by the archimedian property).
By what you proved in 1., find $\delta_{N} > 0$ so that $(x - \delta_{N}, x + \delta_{N})$ contains rational numbers only with denominators larger than $N$.
Then if $y \in (x-\delta_{N}, x + \delta_{N})$, $y$ could either be rational or irrational. If $y$ is irrational, we have $|f(x) - f(y)| = |0 - 0| = 0 < \epsilon$ (since $f(z) = 0$ if $z$ is irrational).
On the other hand, if $y$ is rational, we have $y = p/q$ with $q \geq N$. So $|f(x) - f(y)| = |0 - f(y)| = |f(y)| = |1/q| \leq |1/N| < \epsilon$, and this is true for every $y \in (x-\delta_{N}, x + \delta_{N})$.
Thus, given any $\epsilon > 0$, if $x \in (0,1)$ is irrational, we found $\delta > 0$ (which was actually $\delta_{N}$ in the proof) so that $|x - y| < \delta$ implies $|f(x) - f(y)| < \epsilon$.
Best Answer
Note that your limit is $0$ whenever $$ \left |\sin\left(\frac{\pi x}{2}\right)\right|<1 $$ and $1$ if $\sin\left(\frac{\pi x}{2}\right)=\pm 1$.
So you have discontinuities whenever $$ \sin\left(\frac{\pi x}{2}\right)=\pm 1 $$ well this occurs whenever $$ \frac{\pi x}{2}=\pi k $$ for any integer $k$. Or when $$ x=2k $$ an even integer. These are the points of discontinuity, since your function is $$ f(x)=\begin{cases}1&x=2k,\;k\in \mathbb{Z}\\ 0&\text{otherwise}\end{cases} $$