If $\displaystyle \lim_{n\to\infty} a_n=L$ then prove using the limit definition that: $\displaystyle \lim_{n\to\infty} -2a_n=-2L$.
From the given and the definition we know that: $L-\epsilon<a_n<L+\epsilon$ multiply that by $-2$ and we get: $2\epsilon-2L>-2a_n>-2L-2\epsilon\Rightarrow |-2a_n+2L|<2\epsilon$ which concludes that: $\displaystyle \lim_{n\to\infty} -2a_n=-2L$.
I feel like I was cheating by doing that multiplication by $-2$, is it alright? it's still true for $2\epsilon$ right?
The general idea of you solution is very much correct. The only thing that is missing is that we have: $L-\epsilon<a_n<L+\epsilon$, for sufficiently large $n$.