Closure of regular languages under non-contiguous subsequence
Here are several ways of showing this:


Starting with a DFA/NFA for $L$, add $\epsilon$ transitions parallel to all other transitions.
Apply the regular substitution that maps $\sigma \in \Sigma$ to $\{\sigma,\epsilon\}$.
Starting with a regular expression for $L$, replace all copies of each $\sigma \in \Sigma^*$ by $\epsilon + \sigma$.