Complete convergence is equivalent to convergence a.s. under independence
Your idea is fine and there is small step to finish it. I write the proof anyway.
Suppose that you do have the a.s. convergence but not the complete convergence. Therefore you have: $$ \sum_{n=1}^{\infty} P(\mid X_n-X\mid >\epsilon)=\infty \space\exists\epsilon > 0 $$ For such $\epsilon$, you can use second Borel-Cantelli to prove that: $$ P(\limsup_{n\rightarrow\infty} \mid X_n-X\mid > \epsilon)=1 $$ But from a.s. convergence you have $\displaystyle P(\lim_{n\rightarrow\infty} X_n=X)=1$ which means that for all $\epsilon>0$, $\displaystyle P(\limsup_{n\rightarrow\infty} \mid X_n-X\mid > \epsilon)=0$ which is a contradiction. Therefore $X_n$ should have complete convergence.