Is this limit proof right? 
If $\displaylines{\lim_{n\rightarrow \infty}a_n=a}$ then $\displaylines{\lim_{n\rightarrow \infty}\alpha a_n=\alpha a}$.

Proof:
Let $\epsilon>0$ and $\alpha \neq 0$. Then for all $(\frac{\epsilon}{|\alpha|}>0)(\exists n(\epsilon)\in \mathbb{N})$ so that for all $n>n(\epsilon)$ we have $|a_n-a|<\frac{\epsilon}{|\alpha|}$.
Take $n(\epsilon)=\frac{\epsilon}{|\alpha|}$. Now,
$|\alpha a_n - \alpha a|=|\alpha||a_n-a|<|\alpha|\frac{\epsilon}{|\alpha|}=\epsilon$.

You have the right idea. Why did you take $n(\epsilon)=\frac\epsilon{|\alpha|}$, and where was that used? It should be the same $n(\epsilon)$ that was used in the first limit. It should be more along the lines of:
For all $\epsilon>0$, there exists $n(\epsilon)>0$ such that if $n>n(\epsilon)$, then $|a_n-a|<\frac\epsilon{|\alpha|}$, and therefore,
$$|\alpha a_n-\alpha a|=|\alpha||a_n-a|<|\alpha|\frac\epsilon{|\alpha|}=\epsilon.$$
