How to Determine Existence of Turing Reducible Languages? Are finite and recursive instances of $L$ possible with the following constraints?

$\text{L $\subseteq$ {0,1}*}$ and $L \leq HALT \leq \overline L$ where $\overline L = \{\text{ x $\in$ {0,1}* : x $\notin$ L }\}$.  

NOTE: Raphael and Yuval kindly pointed out that I was overlooking $\overline L$ is undecidable. Aside from this oversight, are there any other issues, e.g. conceptual misunderstandings, logical fallacies, etc. in the explanation below? 

As I understand it:


A finite instance of $L$ is possible. Since the language is finite, an evaluating Turing Machine (TM) could not loop indefinitely on input, and consequently would halt either in an "accept" or "reject" state. Since the language stops on all inputs, it is recursive and thus can be reduced to $HALT$, which is recursively enumerable. 
Since a finite instance (which is recursive) is possible, logically a recursive $L$ must also be possible.
To narrow down the search space, consider what you know about closure properties and HPL. 


The class of decidable languages $R$ is closed against complement.
HPL is not in $R$.


You can conclude that $L \not\in R$;¹ otherwise, $\overline{L} \in R$ and therefore $\mathrm{HPL} \not\leq_T \overline{L}$.

On the other end, you need that $L \in RE$ (the set of semi-decidable languages) -- otherwise $L \not\leq_T \mathrm{HPL}$.

In combination, you need $L \in RE \setminus R$ which answers your question².




In particular, $L$ can not be finite since all such languages are decidable. The decider just checks a hard-coded table of all words in the language. With the same table you can also decide the complement.
Note how this implies $\overline{L} \not\in R$ which is enough for $HPL \leq_T \overline{L}$.