$f: \mathbb{R}\rightarrow\mathbb{R}$, prove that if $f(x)$ is continuous but not uniformly continuous on $\mathbb{R}$, then neither is $\sin f(x)$

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a function, if $f(x)$ is continuous on $\mathbb{R}$ but is not uniformly continuous on $\mathbb{R}$, then prove that $\sin f(x)$ is not uniformly continuous on $\mathbb{R}$. What I have worked out: That $f(x)$ is not uniformly continuous on $\mathbb{R}$ implies that $\exists\varepsilon>0, \forall \delta>0,\exists x_1,x_2$ $s.t.$ $|x_1-x_2|<\delta$ but $f(x_1)-f(x_2)>\varepsilon$. By the continuity of $f$ we can infer that $\forall\varepsilon'\in(0,\varepsilon],\exists x_2,x_3 $$s.t. |x_2-x_3|<\delta$ but $f(x_3)-f(x_2)=\varepsilon'$. Hence it suffices to show that given $d>0$, $\sin[a,a+d] :=\{\sin x|a\leq x\leq a+d\}$ which is an interval has a length which is always no less than $m$ for some $m>0$ fixed as $a$ varies. Although this is intuitive by $\sin x$'s diagram, I don't know how to prove it in a clear way.

Prove by contradiction. Suppose there exists a sequence $(a_n)$ such that the length of $\{\sin x: a _n \leq x \leq a_n+d\}$ tends to $0$. Replacing $a_n$ by $a_n+2k_n \pi$ for a suitable integer $k_n$ we may suppose $(a_n) \subset [0,2\pi]$. So $(a_n) $ converges along some subsequence to a real number $a$. Now it is easy to check that the length of $\{\sin x: a+\epsilon \leq x \leq a+d-\epsilon\}$ is $0$ for $0<\epsilon <\frac d 2$. But this is impossible since $\sin x$ is not constant on any interval.