I claim that a subframe of even a finite Isbell Hausdorff frame is not necessarily Isbell Hausdorff. Furthermore, I claim that a finite frame is Isbell Hausdorff if and only if it is a Boolean algebra. In [1][Thm 2.3], Isbell states that every regular frame is strongly Hausdorff. In particular, since every complete Boolean algebra is a regular frame, every complete Boolean algebra is strongly Hausdorff. On the other hand, again in [1][Thm 2.3], Isbell states that the space of points in every strongly Hausdorff frame is a Hausdorff space. In particular, if $D$ is a strongly Hausdorff finite distributive lattice, then the space of points $Sp(D)$ of $D$ is a finite Hausdorff space, so $Sp(D)$ is a discrete space. Therefore the frame $\Sigma(Sp(D))$ of open subsets of $Sp(D)$ is a Boolean algebra. However, every finite distributive lattice is clearly a spatial frame, so $\Sigma(Sp(D))\simeq D$, so $D$ is a Boolean algebra.

It should note that for infinite frames, the action of taking a subframe of a frame preserves no separation axioms whatsoever. In [2. p.53], it is shown that every frame is isomorphic to a subframe of a complete Boolean algebra. Furthermore, complete Boolean algebras satisfy the strongest separation axioms since they are regular, normal, and even paracompact, and they satisfy strong zero-dimensionality axioms. In fact, complete Boolean algebras are ultraparacompact. And of course, the topologies are precisely the subframes of the power sets $P(X)$. On the other hand, a sublocale of a strongly Hausdorff locale is always strongly Hausdorff, so sublocales behave radically differently than subframes.

[1] Isbell, John R. Atomless parts of spaces. Math. Scand. 31 (1972), 5–32.

[2] Johnstone, P. T. Stone Spaces. Cambridge Cambridgeshire: Cambridge UP, 1982.