As @Adam noted, a function is Riemann integrable if and only if it is almost everywhere continuous. Giving a proof in terms of the Darboux integral (upper and lower sums) would also be rigorous. However, all of these might be a bit mysterious to a high school calculus student. The main pedagogical point to make with a problem like this would be along the lines of noting that not every function is integrable and that we need to pay attention to hypotheses and definitions.
Personally, I would take the following approach when trying to explain intuitively why this function isn't Riemann integrable—citing a theorem from measure theory feels a bit unsatisfying. Basically, if we're going to move into the weird word of measure theory, let's start by just dipping in a toe (the indicator function on the rationals), gaining some intuition, and then building towards the function given in the problem.
Show that between any two rational numbers there is an irrational number and vice versa. If you're comfortable with measure theory, you could discuss the measures of the rationals and irrationals.
Analyze the indicator function of the rationals:
$$\chi_\mathbb{Q} (x) := \begin{cases} 1 & x \in \mathbb{Q}, \\
0 & x \in \mathbb{R}\setminus \mathbb{Q}. \end{cases}$$
Discuss that it is indeed a well-defined function, that it is nowhere continuous, and is not Riemann integrable. The numbers here are much easier to deal with than the original function! You could even show how to create a sequence of Riemann sums on the interval $[0,1]$ such that the limit is any number between $0$ and $1$.
Repeat for the function $g(x) = x\cdot \chi_\mathbb{Q}(x)$. Be sure to note that this function is continuous at only a single point! Alternatively, see if they can come up with this one on their own--ask for a well-defined function $g:\mathbb{R} \to \mathbb{R}$ that is continuous at only a single point.
Finally we can analyze the given function $f(x) = x+ x\cdot\chi_\mathbb{Q}(x)$, observing that it has many of the same properties as the previous two functions.
As a coda, note that "infinitely many disconintuities" is not the same concept as "almost everywhere (dis)continuous". The canonical example would be Thomae's function, defined by
$$f(x) := \begin{cases} \frac{1}{q} & x=\frac{p}{q}\in\mathbb{Q} \text{ in lowest terms, with $p \in \mathbb{Z}$ and $q \in \mathbb{N}$,} \\ 0 & x \in \mathbb{R}\setminus \mathbb{Q}. \end{cases}.$$
This function has the following curious properties:
- It is indeed a well-defined function.
- It is discontinuous at every rational number and continuous at every irrational number (straightforward $\epsilon$-$\delta$ proof).
- It is Riemann integrable! It's easiest to just draw boxes, but a rigorous proof is reasonably straightforward as well. Intuitively, there are only so many spikes in the function over any given height and these get contained in narrower and narrower rectangles.
If you introduce Thomae's function, you will invariably be asked if there is a function that is instead continuous at all the rationals and discontinuous at all the irrationals. It can be worth having them think about it for a few days, but the answer is no as this would contradict the Baire Category Theorem.