Informally speaking, you have to show that the terms of the interleaved sequence get closer and closer to $L$ as the index becomes larger and larger. Each of the sequences $(A_n)$ and $(B_N)$ has this property; it seems obvious that the interleaved sequence will have this property.
But that's all rather vague. Formally, you need to show that for any given $\epsilon>0$, there is a positive integer $N$ so that whenever $n>N$, the $n^\text{th}$ term of the interleaved sequence, call it $I_n$, satisfies $|I_n-L|<\epsilon$.
Towards that end, here's a hint:
Let $\epsilon>0$. Choose $N_1$ so that for $n>N_1$, we have $|A_n−L|<\epsilon$. This can be done since $(A_n)$ converges to $L$. Then choose $N_2$ so that for $n>N_2$, we have $|B_n−L|<\epsilon$. This can be done since $(B_n)$ converges to $L$.
Now, what can you say about the $n^{\rm th}$ term of the interleaved sequence if $n>2\max\{N_1,N_2\}$?