Yes, if $K$ is imperfect then any finite extension $L$ of $K$ is also imperfect.
Indeed if $char.K=p$ and $a\in K$ has no $p$-th root in $K$, then any perfect field $P$ containing $K$ must contain all $a^{p^{-n}} \: (n\geq 1)$ and, since $[K(a^{p^{-n}}):K]=p^n$, the field $P$ must be of infinite dimension over $K$ (since its dimension is $\geq p^n$ for all $n$).
The fact that $L$ is separable or Galois over $K$ will not help: I'm sorry to say that $L$ will still be imperfect (but aren't we all...)