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 [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 [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.
 Isbell, John R. Atomless parts of spaces. Math. Scand. 31 (1972), 5–32.
 Johnstone, P. T. Stone Spaces. Cambridge Cambridgeshire: Cambridge UP, 1982.