Suppose $\left\{ x_{n}\right\} $ is convergent. Prove that if $c\in\mathbb{R} $, then $\left\{ cx_{n}\right\} $ also converges. Good morning! I wrote a proof of the following exercise but I don't know if it is fine:
Suppose $\left\{ x_{n}\right\} $ is convergent. Prove that if $c\in\mathbb{R} $, then $\left\{ cx_{n}\right\} $ also converges.
Proof:
If $\left\{ x_{n}\right\} $ converges, then by definition
$$\lim_{n\rightarrow\infty}x_{n}=x.$$
For $\frac{\epsilon}{\mid c\mid}>0$, let $N\in\mathbb{\mathbb{N}}$ be such that if $n>N$, then $\mid x_{n}-x\mid<\frac{\epsilon}{\mid c\mid} $.
Now, we have this:
$$|cx_{n}-cx|=|c(x_{n}-x)|=|c|| x_{n}-x|=|c||x_{n}-x|<|c|\frac{\epsilon}{|c|}=\epsilon\Rightarrow c\lim_{n\rightarrow\infty}x_{n}=cx$$ and $\left\{ cx_{n}\right\} $ converges.
Please review this...

Looks like you have the meat of the problem, so I have a few smaller details to point out.

$(1)$ Make a special case for $c = 0$. The quantity $\frac{\varepsilon}{\left| c\right|}$ only exists for nonzero values of $c$.
$(2)$ When you say "If $\left\{ x_{n}\right\}$ converges then by definition
  $\lim_{n\rightarrow\infty}x_{n}=x$", I think that is something that could be moved into the problem statement instead of needing to be stated during the proof. For example "Suppose $x_n \to x \ldots$" is a concise way to state that. By no means a big deal, but taking it out of the proof portion will shorten the proof, which is usually ideal.
$(3)$ From here you have to manipulate your inequality to the point that it says $$\left|cx_n -cx \right|< \varepsilon$$ At this point exactly is when you have proven the claim. I do not see that concisely stated in your work; there is a lot of unnecessary inequality manipulation.
$(4)$ Great contribution to MSE! Always glad to see when someone clearly shows their work.

