Let $G = SU(3)$, and let T be the subgroup of G of diagonal matrices.
How can we prove that $\lbrace T \cap gTg^{-1}, g \in G \rbrace = \lbrace T, T_e, T_1 , T_2, ,T_3 \rbrace $?
Where $T_e$ is the set of matrices of the form $\lambda I$, $\lambda \in \mathbb{C}, I $ is the identity matrix.
$T_1$ are diagonal matrices in T s.t the first and second components are equals .
$T_2$ are diagonal matrices in T s.t the second and third components are equals .
$T_3 $ are diagonal matrices in T s.t the first and third components are equals .
$\newcommand{\inv}{^{-1}}$Case 1. Suppose first that $T\cap gTg\inv$ contains a matrix $a$ with three unequal diagonal entries.
Observing that $g\inv a g$ is also diagonal, and that it has the same spectrum as $a$, we see that the diagonal entries of $g\inv a g$ must be a permutaion of the entries of $a$. Therefore there exists a permutation matrix $\sigma $ (i.e. the matrix representing a linear transformation permuting the canonical basis) such that $$ g\inv a g = \sigma \inv a \sigma . \tag1 $$ It follows that $g\sigma\inv$ commutes with $a$, but it is well known that only diagonal matrices have this property. Therefore $g\sigma\inv=d$, for some diagonal matrix $d $, hence $g = d\sigma $, and one may now easily prove that $gTg\inv=T$, so $T\cap gTg\inv=T$.
Case 2. Suppose now that $T\cap gTg\inv$ contains only matrices $a$ with three identical diagonal entries.
If $a\in T\cap gTg\inv$, then we may write $a=\lambda I$, and it is clear that $\lambda ^3=\det(a) =1$, so $$ \lambda \in \Lambda := \{1, e^{2\pi i/3}, e^{4\pi i/3}\}. $$ This shows that $$ T\cap gTg\inv \subseteq \Lambda I, $$ and the reverse inclusion is obviously true (regardless of $g$).
Case 3. Suppose now that $T\cap gTg\inv$ does not contain any matrix with three unequal diagonal entries, but that it contains a matrix $a$ with two unequal diagonal entries. Without loss of generality we may assume that $$ a=\pmatrix{\lambda & 0 & 0\cr 0 & \lambda & 0\cr 0 & 0 & \mu } $$ where $\lambda \neq \mu $ (otherwise we may conjugate everything by a cyclic permutation matrix).
Just like case (1), we may find a permutation matrix $\sigma $ satisfying (1), so that $g\sigma\inv$ commutes with $a$. The commutator of $a$ in $SU(3)$ may be shown to coincide with the set of matrices of the form $$ v=\pmatrix{u_{11} & u_{12} & 0\cr u_{21} & u_{22} & 0\cr 0 & 0 & \tau }, $$ where $$ u:=\pmatrix{u_{11} & u_{12} \cr u_{21} & u_{22}}\in U(2), $$ and $\tau =\det(u)\inv$. So we deduce that there is some $v$, as above, such that $g\sigma\inv=v$, whence $g=v\sigma $.
With a little more effort one may now show that $T\cap gTg\inv$ coincides with $T_1$ (as defined in the original post).
The other cases correspond to alternatives to case (3), depending on which of the two diagonal entries of $a$ agree, and they lead to $T\cap gTg\inv$ coinciding with $T_2$ or $T_3$.
This shows that $$ \lbrace T \cap gTg^{-1}: g \in G \rbrace \subseteq \lbrace T, T_e, T_1 , T_2, ,T_3 \rbrace, $$ and the reverse inclusion may be easily shown by choosing $g$ of the apropriate form, as suggested by the above cases.