The factor group $\mathbb{R} / \mathbb{Z}$ under addition has an infinite number of elements of order 4.
The answer is false, but I cannot still understand it. I thought that it is true since there are infinite number of real numbers $x$ such that $4x \in \mathbb{Z}$. Why that this argument is incorrect?
Your argument is incorrect, because the problem is not about counting elements of $\mathbb R$.
It is instead about counting elements of $\mathbb R / \mathbb Z$. In other words, you should take the set $$\{x \in \mathbb R \mid 4x \in \mathbb Z\} $$ then you should subdivide it into cosets modulo $\mathbb Z$, and then you should count the number of cosets. That will be, almost by definition, the number of elements of the factor group $\mathbb R / \mathbb Z$ such that their order divides $4$ (then you'll have to winnow out the elements of order $2$ and $1$).
For example, the set $\frac{1}{4} + \mathbb Z = \{... \,-\frac{7}{4}, \, -\frac{3}{4}, \, \frac{1}{4}, \, \frac{5}{4}, \, \frac{9}{4} \,...\}$ consists of infinitely many real numbers $x$ such that $4x \in \mathbb Z$, but this set is just a single coset modulo $\mathbb Z$. It is just a single element of $\mathbb R / \mathbb Z$ of order dividing $4$, and in fact its order is exactly $4$ because $2 \cdot \frac{1}{4} \not\in \mathbb Z$ and $1 \cdot \frac{1}{4} \not\in \mathbb Z$.