Show that $\sin(f)$ is a continuous mapping in $\mathcal{B}(M)$ 
Show that $\sin:\mathcal{B}(M)\to \mathcal{B}(M)$ is a continuous mapping, where $\mathcal{B}(M)$ is the space of bounded real valued functions equipped with the $\Vert\cdot\Vert_{\infty}$ sup norm.

My approach:
We already know that $h:\mathbb{R}\to\mathbb{R}$ where $h:=\sin(x)$ is a uniformly continuous function. So for an arbitrary $\epsilon>0$ we find a $\delta>0$ such that for all $x,y\in\mathbb{R}$ we have $|x-y|<\delta\implies |f(x)-f(y)|<\epsilon$.
Now we take two $f,g\in\mathcal{B}(M)$ such that $\Vert f-g\Vert_{\infty}<\delta$. This implies for all $x\in M$ that $|f(x)-g(x)|<\delta\implies |\sin(f(x))-\sin(g(x))|<\epsilon\implies \Vert \sin(f)-\sin(g)\Vert_{\infty}<\epsilon$.

Is this correct? I am a bit skeptical because it seems that we don't need the boundedness.
 A: The OP and I are debating a subtle point, too subtle for the tiny space of comments.
His solution of the problem is fine, but missing a necessary remark at least.  This "answer" just addresses that point.

*

*For any bounded function $f:[0,1]\to \mathbb R$ define $\|f\|_\infty=\sup\{|f(x)|:x\in [0,1]\}$.


*Define $D[0,1]$ to be the linear  space of all bounded derivatives $f:[0,1]\to \mathbb R$ equipped with the norm $\|f\|_\infty$.  This is a normed linear space.  In fact it is a much-studied Banach space.  See reference [1].


*Let  $F:\mathbb R\to \mathbb R$  be a uniformly continuous function and  write  $T_F$ for the mapping $f\to F\circ f$,  i..e., this takes any real-valued function $f(x)$ and yields the function $F(f(x))$.
Prove the false statement that $T_F$ is a continuous map from the space $D[0,1]$ to itself.
Copying from the OP:

We already know that $F:R→R$   is a uniformly continuous function. So
for an arbitrary $ϵ>0$ we find a $δ>0$ such that for all $s,t∈R$ we
have $|s-t|<δ⟹|F(s)−F(t)|<ϵ.$
Now we take two $f,g∈D[0,1]$  such that $ ∥f−g∥_∞<δ$. This implies for all
$x∈[0,1]$ that $|f(x)−g(x)|<δ⟹|F(f(x))−F(g(x))|<ϵ⟹∥T_F(f)−T_F(g)∥_∞<ϵ$.
Thus $T_F$ is a continuous map.

Spoiler:  One can give an example of a function $f\in D[0,1]$ such that $f^2$ is not in $D[0,1]$.  In fact given any continuous, increasing function $F$  that is not linear there exists a bounded derivative $f$ such that $F\circ f$ is not a derivative.

REFERENCE:
[1] https://www.jstor.org/stable/44151124
