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 $x\in(0,1)$. $\Box$ 
 A: 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 $0 < f(x) < 2$.
A: Your proof works. Here is another one.
It is easily seen that $\lim_{x\to 0} f(x) = 1$. Also, $f$ is defined at $x=1$. Hence, we can extend $f$ by continuity to the function $g: [0,1] \to \mathbb R$,
$$g(x) = \begin{cases} 1, \text{ if $x = 0$} \\ f(x), \text{ if $x \in (0,1]$} \end{cases}$$
A continuous function on a compact set is bounded. Hence, $g$ is bounded and therefore so is $f$.
A: You make an error that $\sin x + 1 > M$ means $\sin x > M -1$.  I say let $M \ge 2$ in the beginning.  Or better yet...
Since are are trying to prove something is bounded, rather than proving by contradiction from assuming $|f(x)| > M$ for all $M$ and finding it isn't true for some $M$, simply do it directly instead.
If $|\frac {(x+1)\sin x}x| = |f(x)|$ then
$(x+1)\sin x = x\sin x + \sin x = |f(x)|x$
$x\sin x + x > x\sin x + \sin x = |f(x)|x$
$\sin x + 1 > |f(x)|$
$\sin x > |f(x)|-1$ 
$1 \ge \sin x > |f(x)|-1$ so
$2 > |f(x)|$.
So $f(x)$ is bounded.
