I think I have a solution (thanks to Guido de Philippis).
First I want to show that if $x>0$ is sufficiently small for all positive integers $n$ one has: $$ \sin x \le n \sin \frac x n \le \tan x. $$ To prove this let $y=\frac x n$ so that our thesis becomes $$ \sin ny \le n \sin y \le \tan ny. $$ We are going to use induction on $n$.
For the first inequality we have $$ \sin (n+1)y = \sin ny\cdot \cos y + \cos ny \cdot \sin y \le \sin ny + \sin y. $$ Using inductive assumption we obtain $$ \sin(n+1)y \le n\sin y + \sin y = (n+1)\sin y $$ which is what we wanted to prove.
For the second inequality we have: $$ \tan (n+1)y = \frac{\tan ny + \tan y}{1 - \tan ny \cdot \tan y} \ge \tan ny + \tan y \ge \tan ny + \sin y $$ supposing that $0\le \tan y$, $0\le \tan ny$, $0\le\tan ny \cdot \tan y < 1$. So by inductive assumption we have $$ \tan (n+1)y \le n \sin y + \sin y = (n+1)\sin y $$ as we wanted to prove.
So we have proved that fixed a sufficiently small positive $x$ one has $$ \frac{\sin x}{x} \le \frac{\sin \frac x n}{\frac x n } \le \frac{\tan x}{x}. $$ The sequence $$ a_n = \frac{\sin \frac x n }{\frac x n} $$ is increasing (see the answer by Iosif Pinelis). Hence $a_n\to \ell$ with $0 < \ell < +\infty$.
Now given any $y<x$ one can find $n=n(y)$ such that $\frac{x}{n+1} \le y \le \frac{x}{n}$ whence $$ a_{n+1} \cdot \frac{n}{n+1} = \frac{\sin \frac x {n+1}}{\frac x n } \le \frac{\sin y}{y} \le \frac{\sin \frac x n}{\frac x {n+1}} = a_n \cdot \frac{n+1}{n} $$ and now if $y\to 0^+$ one has $n\to +\infty$ hence $\frac{\sin y}{y}\to \ell$.