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I'm not able to mathematically prove that the equation
$$x(k+1)=\sin(x(k))$$
converges asymptotically to zero.
By a simple thought it can be concluded that for any $x(0) \in \mathbb{R}$ it applies
$x(1)=\sin(x(0)) \qquad \qquad \quad x(1) \in [-1,1]$
$x(2)=\sin(\sin(x(0))) \qquad \, \, \, \, \, x(2) \in [-0.84,0.84]$
$x(3)=\sin(\sin(\sin(x(0)))) \, \, \, \, x(3) \in [-0.74,0.74]$
...
Which means that with every step the solution is closer and closer to zero, however I'm not able to prove it.
Does anyone know how to solve this problem please?

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This is a great exercise in learning how to apply the tools you learn in your calculus course.

I'm going to limit myself to $x(1) \geq 0$: the case $x(1) \leq 0$ is similar.

Your first observation: that $x(n)$ is getting closer to zero. You have an intuitive idea that $\sin t \leq t$ whenever $t \geq 0$, correct?

And why do you feel this? If your answer sounds like "$\sin t$ simply doesn't grow as fast as $t$", that's a huge clue: use derivatives to prove it.

Now, you get to use another major tool: you have a sequence that is monotone and bounded. Therefore, it has a limit.

Finally, now that you know the limit exists you get to use the calculus of limits to obtain

$$ L = \lim_{n \to +\infty} x(n) $$

How to obtain this limit? Well, the "only" thing you know about $x(n)$ is that $x(n) = \sin(x(n-1))$: what happens if you plug that in?

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