$\mathcal{T}$ is the standard topology and $\mathcal{T}'$ is any other different topology, both on the unit interval $[0, 1]$. Then if $\mathcal{T}' \subsetneq \mathcal{T}$, $[0,1]$ with $\mathcal{T}'$ is not Hausdorff, and if $\mathcal{T} \subsetneq \mathcal{T}'$, $[0,1]$ with $\mathcal{T}'$ is not compact.
For the first case we have to show that $[0,1]$, with any coarser topology than the subspace topology, has two distinct points where you cannot have two disjoint neighborhoods between them. That means we have to remove some open sets from the subspace topology until we get another valid topology, but I don't know in general which open sets I can remove from the subspace topology to yield another topology. But there might even be an infinite number of different topologies, and I have to show that every single one has two distinct points where you cannot have two disjoint neighborhoods between them, which seems absurd, since I don't know how to represent a general form of a topology that represents all topologies smaller than the subspace one.
For the second case, I have to show that in any finer topology than the subspace topology, there exists an open cover of $[0,1]$ which doesn't have a finite subcover. But again I run into a similar issue as the first case.