My attempt: if $(a\_n)^2$ converges, then it converges to a finite limit, then by the definition of the limit we get that : $|(a\_n)^2 - L| < \epsilon $   
 then I opened up the absolute value and got: $L-\epsilon < a\_n^2 < L + \epsilon $. (I really wanted to take a square root and end it here but if any sides of the inequation has a negative value then my proof is false, I tried to come up with a counter example and I couldn't find one too, but I think here I can choose the $\epsilon $ as I like since the limit exists so if I took $\epsilon = \frac {L} 2$, I would have all positive and be able to take square root on all the inequality sides.   
 I wanna make sure if my proof is correct, and if it is, then why choosing one specific $\epsilon $, holds for a proof.   
 if it's not any explanation would be pretty much appreciated.   
 Thanks in advance.