Does an unbounded open subset of $\mathbb{R}$ contain infinitely many multiples of some number?

You can prove this using the Baire category theorem. Let's let $U$ be an unbounded subset of the positive reals, noting that this change does not affect the theorem. Define $U_n = U \cap (n,\infty)$ and define $$V_n = \{x\in\mathbb R^+ : \text{There exists an integer }k\text{ so that }kx\in U_n\}.$$ This set is open as we could also write $$V_n=\bigcup_{k=1}^{\infty}\frac{1}kU_n$$ where each of the sets in the union is open.

We note that $V_n$ is dense in $[0,\infty)$ because its complement cannot contain any intervals - note that if an interval $(a,a+\varepsilon)$ was in the complement of $V_n$, then, $U_n$ must be disjoint from every interval of the form $(ka, ka+k\varepsilon)$. However, the union $\bigcup_{k=1}^{\infty}(ka,ka+k\varepsilon)$ has a bounded complement in $\mathbb R^+$ because for any $k$ such that $k\varepsilon > a$, the interval $(ka,ka+k\varepsilon)$ overlaps with $((k+1)a, (k+1)a+(k+1)\varepsilon)$ and thus this union contains all of $(ka,\infty)$. Since $U$ is unbounded, this cannot be.

Note that the set of numbers for which there are infinitely many multiples in $U$ is exactly $$\mathscr V=\bigcap_{n=1}^{\infty}V_n$$ since if there were only finitely many multiples of $x$ in $U$, then there would be none in some $U_n$. This is a countable intersection of open dense sets in $[0,\infty)$. By the Baire category theorem, $\mathscr V$ is dense in $[0,\infty)$ - and, in particular, it is not empty.

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...)

This develops an idea posted in an answer by @lulu, since deleted (before I could grasp the details, so @lulu is not responsible for my possible misinterpretation :).
I am confused, so let me write a sketch of it before I forget it.

Assume $\sup U=\infty$. Define $a_1<b_1<a_2<b_2<\cdots$ and $n_1,n_2,\dots$ such that:

(i) $[a_k,b_k]\subset U$,
(ii) $\bigl(n_k\cdot(a_k,b_k)\bigr)\cap U\neq\emptyset$,
(iii) $[a_{k+1},b_{k+1}]\subset\bigl(n_k\cdot(a_k,b_k)\bigr)\cap U$.

Then $(a_k,b_k)\cap\frac U{n_k}\neq\emptyset$ and $\frac{[a_{k+1},b_{k+1}]}{n_k}\subset(a_k,b_k)\cap\frac U{n_k}\subset[a_k,b_k]\subset U$.

Moreover, $\frac{[a_{k+2},b_{k+2}]}{n_k n_{k+1}}\subset\frac{(a_{k+1},b_{k+1})\cap U}{n_k}\subset\frac{[a_{k+1},b_{k+1}]}{n_k}\subset U$.

Thus $\frac{[a_{k+2},b_{k+2}]}{n_k n_{k+1}}\subset\frac{[a_{k+1},b_{k+1}]}{n_k}\subset [a_k,b_k]\subset U$.

The idea is to "diagonalize" and get a nested sequence of closed intervals, $[a_1,b_1]\supseteq\frac{[a_2,b_2]}{n_1}\supseteq\frac{[a_3,b_3]}{n_1 n_2}\supseteq\cdots\supseteq\frac{[a_{1+j},b_{1+j}]}{n_1 n_1\cdots n_{j}}$. (Then take $a$ in their intersection.) I have to think more, but I will post this.