Let me rephrase the answer in different terms.

Since $H_i$ are idempotent and self-conjugate, they are orthogonal projectors onto some subspaces $W_i$. Set $U=W_1\cap W_2$, and let $U_i$ be the orthogonal complement of $U$ in $W_i$. Finally, let $U^\perp$ be the orthogonal complment of $W_1+W_2$ in the whole space $V$. Then $V=U\oplus U_1\oplus U_2\oplus U^\perp$ (but $U_i$ are not necessarily orthogonal to each other). Notice here that $H_i(U_{3-i})\subseteq U_i$ by orthogoality.

Now, each vector $v$ is uniquely expanded as $v=u+u_1+u_2+u^\perp$ with each summand lying in the corresponding space. If $v$ is an eigenvector of $H_\mu$ with the eigenvalue $\mu$, then
$$
  \mu u+\mu u_1+\mu u_2+\mu u^\perp
  =H_\mu v
  =u+\mu u_1+(1-\mu)H_2(u_1)+(1-\mu)u_2+\mu H_1(u_2)
  =u+\mu(u_1+H_1(u_2))+(1-\mu)(u_2+H_2(u_1))+0,
$$
the terms in the right-hand part also lie in the corresponding subspaces. Hence $u=u^\perp=0$, $H_1(u_2)=0$ (so $u_2\perp U_1$), and $(2\mu-1)u_2=(1-\mu)H_2(u_1)$. The last equality cannot hold unless $u_2=0$ or $\mu=1/2$, since $u_2\perp U_1$. So $v=u_1\in \mathop{\rm Im} H_1$ and $H_2(u_1)=0$ (thus $u_1\in\mathop{\rm Ker} H_2$), as required.

Perhaps, this language is better for generalizations?