Prove that $f(x) = \frac{(x+1)\sin x}{x}$ is bounded at $(0,1)$. Is it a valid proof? Proof:
We assume that $f(x)$ is unbounded at $(0, 1)$. So for any $M\in \mathbb{R}$
there exists an $x\in (0,1)$ such that:  
$|f(x)| = |\frac{(x+1)\sin x}{x}| > M \Rightarrow^{(1)} \frac{(x+1)\sin x}{x}>M
\Rightarrow x\sin x+\sin x > Mx \Rightarrow^{(2)} x(\sin x + 1)>Mx \Rightarrow
\sin x+1 > M \Rightarrow \sin x>M-1.$  
(1).$\sin x > 0$ for any $x\in(0,1)$
(2).$x>\sin x$ for $x\in(0,1)$
We can choose $M=3$, so:
$\sin x>2.$  
And this is a contradiction because $\sin x \leq 1$ for any $x\in(0,1)$. $\Box$ 

Here's a direct proof, similar to yours, but perhaps a little simpler . . .

On the interval $(0,1)$, we have $0 < \sin x < x$, hence
$$0 < \frac{(x+1)\sin x}{x} < \frac{(x+1)\,x}{x} = x+1 < 1 + 1 = 2$$
so on the interval $(0,1)$, we have $0 < f(x) < 2$.
