Question: I have decided to assume that $a<0$ and let $\epsilon=-a>0$. Then I want to come to a contradiction.

$|a\_n-a|<-a$.

Since we have $a\_n>a$ from our assumption,

$a\_n-a<-a$

so $a\_n<0$ which is a contradiction.

Answer: 
If $a<0$ then since $a\_n\to a$, there is an integer $N$ such that whenever $n>N$, we must have $a\_n\in (a,0)$, i.e. $a\_n<0.$ And this is a contradiction. 

In fact, the same reasoning shows that if $a\_n>0$ for all $n$, then $any$ limit point $a$ of $(a\_n)$ must satisfy $a\ge 0.$
This proof has a flaw since there is no restriction that $a\_n>a$ (consider $a\_n=(-{1\over 2})^n+1$ and $a=1$) but it can be simply fixed. Just conclude from $|a\_n-a|<-a$ that $$a<a\_n-a<-a$$ or equivalently $$2a<a\_n<0$$ which is **now** a contradiction. Except that, your proof is fine.