Does an unbounded open subset of $\mathbb{R}$ contain infinitely many multiples of some number? If $U$ is an unbounded open subset of $\mathbb{R}$, does there exist some $a$ such that infinitely many integer multiples of $a$ belong to U?

Let $I_1 = (a,b)$ be some interval of positive numbers contained in the set $S$. Consider the union of its multiples, save for $I_1$ itself: $$U_1 := \bigcup_{n = 2}^{\infty} n \cdot I_1.$$ You can check that there exist an $N_1$ so that for all $x \geq N_1$, $x \in U_1$. As $S$ is unbounded, there must exist an $x_1 \in S \cap U_1$. Therefore, $x_1$ will be element of some $n_1 \cdot I_1$. Let $I_2$ be an interval containing $x_1$ in $I_2 \subset (n_1 \cdot I_1) \cap S$. All numbers in $I_2$ are multiples by $n_1$ of numbers in $I_1$. 
Define $$U_2 := \bigcup_{n = 2}^{\infty} n \cdot I_2.$$ Again there exists an $N_2$ so that for all $x \geq N_2$, $x \in U_2$. As $S$ is unbounded, there must exist an $x_2 \in S \cap U_2$. $x_2$ will be element of some $n_2 \cdot I_2$. Let $I_3$ be an interval containing $x_2$ and subset of  $I_3 \subset (n_2 \cdot I_2) \cap S$. All numbers in $I_3$ are multiples by $n_2$ of numbers in $I_2$, and therefore multiples by $n_1 \cdot n_2$ of numbers in $I_1$. 
Continue this construction inductively, so that all $I_n$ consist of multiples of all intervals preceding them in the sequence.
We now reduce these intervals $I_n$ to compact, closed intervals $J_n$ by $(a, b) \mapsto [a + \frac{(b-a)}{3}, b - \frac{(b-a)}{3}]$, i.e. the inner third of the interval. The desired properties from above concerning multiples still hold.
Now, by the axiom of choice, determine a sequence $y_n \in J_n$. Divide it to get a sequence in $J_1$: $$z_n := \frac{y_n}{\prod_{i = 1}^{n - 1} n_i}.$$
As it is a sequence in a compact, closed and bounded interval, it has a converging subsequence $\sigma$. Its limit, $Z$, lies in $J_1$, but also in $\frac{1}{n_1} J_2$, in $\frac{1}{n_1 \cdot n_2} J_3$ and so forth, as $\sigma$, save for some initial terms, lies completely in  $\frac{1}{n_1} J_2$, in $\frac{1}{n_1 \cdot n_2} J_3$ and so forth. Therefore, infinitely many multiples of $Z$ lie in $S$.
(I hope someone can come up with a more elegant proof, but brute force works sometimes...)
