Is this limit correct? In $\mathbb R$, Suppose $\{x_n\}\to 0 $ and $\{y_n\}$ is bounded.
In order to prove $\lim_{n\to \infty}x_ny_n=0$, I proved:
$\lim_{n\to \infty}|x_ny_n|\leq M\lim_{n\to\infty}x=0$, and conclude the limit without absolute value exists, Is it correct?

It depends on what machinery (in the way of prior theorems) is available to you. We solve the problem in detail, using only the definition. 
We want to prove that $\lim_{n\to\infty} x_ny_n=0$. So we want to show that for any $\epsilon \gt 0$, there is an $N$ such that if $n\gt N$ then $|x_ny_n -0|\lt \epsilon$. Let $\epsilon \gt 0$ be given.
Since the sequence $\{y_n\}$ is bounded, there exists an $M$ such that $|y_n|\lt M$ for all $n$. We may without loss of generality assume that $M\gt 0$.
Since the sequence $\{x_n\}$ has limit $0$, there is an $N$ such that $|x_n|\lt \dfrac{\epsilon}{M}$ if $n\gt N$.
Thus if $n\gt N$, we have
$$|x_ny_n| \le |x_n| M \lt \dfrac{\epsilon}{M} =\epsilon.$$    
Remark: The basic programs for a computer need to be written in machine language, and can be quite tedious to write.  After a while, instead of going back to machine language each time, we produce a collection of  subroutines that can be called on. The basic limit theorems are such subroutines. Once these are available, we can much of the time call on them, instead of returning to $\epsilon$-$\delta$ or $\epsilon$-$N$ each time. 
Your solution used (correctly) several such results. Suppose that $\lim x_n$ exists and $|y_n|\le M$. It is quite possible that the (correct and intuitively reasonable) result that $\lim|x_ny_n|\le M\lim_{n\to\infty}|x_n|$ has in fact not been proved in your course. In that case, there would be a formal gap in your argument.
