Prove that there is no computability reduction HP $\le$ $\Sigma$*
Let us actually prove more: If $L$ is a language and $L \not\in \mathsf{R}$, then $L \not\le_\mathrm{T} L'$ for any $L' \in \mathsf{R}$. (Here, $\le_\mathrm{T}$ indicates a Turing reduction; this is synonymous with your notion of a "computability" reduction.) In other words, $\mathsf{R}$ is closed under Turing reductions.

Suppose towards a contradiction that such a reduction exists and is computed by a Turing machine $M$. Furthermore, let $M'$ be a decider for $L'$. Then there is a Turing machine which solves $L$, namely by directly simulating $M$, answering all of its oracle queries by simulating $M'$ on them, and, at the end, answering what $M$ does. This contradicts $L \not\in \textsf{R}$.

The case of the halting problem follows from it not being in $\mathsf{R}$ (since it is not even in $\mathsf{RE}$) and, naturally, $\Sigma^\ast \in \mathsf{R}$.

An aside: If you find the idea of classes closed under Turing reductions interesting, I suggest you look up the computation-theoretical notion of Turing degree. $\mathsf{R}$ would be the "base case" (i.e., degree zero) in that context.