Is $L=\{\langle M\rangle\mid L(M)\subseteq HP\}\in coRE$?
We would show that $HP \le L$.

Let $f:\Sigma^*\rightarrow \Sigma^*$ be a function such that for any $x\in \Sigma^*$:
$$f(x)=\langle M_x\rangle$$
while $M_x$ is a TM that accepts $x$ and rejects every other word.

$f$ is computable, as it only requires writing the encoding of a very simple TM.

Now, for any $x\in \Sigma^*$:


$L(M_x)=\{x\}$
If $x\in HP$, then $\{x\}\subseteq HP$, and so $\langle M_x\rangle \in L$.
Otherwise, $x\notin HP$, and then $\{x\}\not\subseteq HP$, and so $\langle M_x\rangle \notin L$.


Therefore, indeed $HP \le L$.

Thus, it must be that $L\notin coRE$, because otherwise the reduction would imply that $HP\in coRE$, which is false.