To prove (dis)continuity, you can begin with the following observation:
Observation: Suppose that $I = (a-\varepsilon,a+\varepsilon)$ is an interval of length $2 \varepsilon$ such that $r_n \not \in I$ for $n \leq N$, for some given $N$. Let $J = (a-\varepsilon/2,a+\varepsilon/2)$ be an interval of length $\varepsilon$ with the same centre as $I$. Then, there exists a constant $C = C_\varepsilon$ such that for any $x \in J$ it holds that $|f(x) - f(a)| < C |x-a| + \frac{4}{2^N}$.
Proof: First, use "infinite triangle inequality":
$$ |f(x) - f(a)| \leq \sum_{n} |s_n(x) - s_n(a)| =
\sum_{n \leq N} |s_n(x) - s_n(a)| + \sum_{n > N} |s_n(x) - s_n(a)| $$
where we define
$$s_n(x) := \begin{cases} \left(\frac{\sin(1/(x-r_n))}{2}\right)^n,& \ x \neq r_n\\ 0, & x = r_n \end{cases}$$
We bound the two sums above separately. For the second sum, it will suffice to notice that $| s_n(y) | \leq 1/2^n$, and hence $|s_n(x) - s_n(a)| \leq 1/2^{n-1}$. Summing over $n = N+1,N+2,\dots$ we find $$ \sum_{n > N} |s_n(x) - s_n(a)| \leq \frac{1}{2^{N-2}}.$$
For the first summand, first note that for $n \leq N$ we have $r_n \not \in I$, and hence $ s_n(x) = \left(\frac{\sin(1/(x-r_n))}{2}\right)^n$ for $x \in I$. In particular, $s_n$ is a differentiable function, and we can compute the derivative:
$$ s_n'(x) = \frac{n}{2^n} \frac{1}{(r_n-x)^2} \cos\frac{1}{x-r_n} \sin^{n-1}\frac{1}{x-r_n}.$$
For $x \in J$ we know that $|r_n-x| > \varepsilon/2$, and clearly $\cos(t)$ and $\sin(t)$ are bounded by $1$ in absolute value. It follows that $|s_n'(x)| \leq \frac{n}{2^n} \frac{4}{\varepsilon^2}$.
By the mean value theorem, you can approximate for $x \in J$ : $$|s_n(x) - s_n(a)| \leq |x-a| \cdot \max_{y \in J} s_n'(y) \leq |x-a| \frac{n}{2^n} \frac{4}{\varepsilon^2}.$$
Summing this over $n < N$, and letting $C_\varepsilon = \frac{4}{\varepsilon^2} \sum_{n=1}^\infty \frac{n}{2^n}$ we get:
$$ \sum_{n \leq N} |s_n(x) - s_n(a)| \leq C |x-a|$$
Thus, putting this together, we get:
$$|f(x) - f(a)| \leq C |x-a| + \frac{4}{2^N},$$
as stated at the beginning. $\square$
Let us see how the claim follows from the above observation.
First, take $a \in [0,1] \setminus \mathbb{Q}$. Let $\eta > 0$; our goal is to find $\delta > 0$ such that whenever $|x -a| <\delta$ then also $|f(x) - f(a)| < \eta$. First, select $N$ sufficiently large that $4/2^N < \eta/2$. Next, take $\varepsilon$ sufficiently small that in $I = (x-\varepsilon,x+\varepsilon)$ there are no $r_n$ with $n \leq N$. If $|x-a|<\varepsilon/2$ we now have by the observation above:
$$ |f(x) - f(a)| < C |x-a| + \eta/2$$
It suffices to pick $\delta$ so that $C \delta < \eta/2$ (and $\delta < \varepsilon/2$), and we are done!
If $a = r_m$ is rational, then we can exploit the reasoning above. A little informally, let $\tilde{f}$ be a function which is defined exactly the same as $f$, with the exception that we skip the $m$-th summand: $\tilde{f}(x) = \sum_{n \neq m} s_n(x)$. Now, the same argument as we just used shows that $\tilde{f}$ is continuous at $a$. However, $f(x) = s_,(x) + \tilde{f}(x)$, and clearly $s_m(x)$ is discontinuous at $a$. Thus, $f(x)$ is also discontinuous at $a$.
Moreover, $f$ has the same type of discontinuity as $s_m$ (adding a continuous function does not alter the type of discontinuity), so we have proved that $f$ has a discontinuity point of the second kind.
