The answer is no. The counterexample came to me a few seconds after hitting the post button (I'm sorry). However, this that the counter example is interesting and seems to be closely related to the non-topology of almost everywhere convergence in $L_1$.
The set of all monotone functions from $[0,1]$ to $\{0,1\}$ is a counter example. It is a compact (sublattice) in each $L_p$, is (a closed sublattice) in $L_∞$ and for every measurable set $E\subseteq [0,1]$, $λ(E)>0$, the set $X_E=\{f\chi_E: f∈X\}$ is not compact in $L_\infty$.
To see this notice for each $g\in\{f\chi_E: f∈X\}$ is separated from $\{f\chi_E: f∈X\}$ by an $L_\infty$-neighbourhood. Thus, in the relative topology of $L_\infty$ the singleton sets in $X_E$ are both open and closed. But $X_E$ has uncountably many elements. So it is not compact in $L_\infty$.