Before I pose my problem I should mention how many times this general question has been posted (and answered) on this site. As you will see, there are some fantastic explanations of the case when the eigenspaces are 1D, and then a statement equivalent to "its more complicated but similar in the case where eigenspaces aren't 1D" is given for the general case.
The set up: $A: V \to V$ and $B : V \to V$ are both symmetric and $AB = BA.$ The claim is that there exists a single eigenbasis such that both $A$ and $B$ are diagonal, or in other words, there exists a $U$ such that $A = UA'U^T$ and $B = UB'U^T,$ where $A', B' $ are diagonal.
The standard procedure goes by showing that $V = V_1 \oplus V_2 \oplus \cdots \oplus V_m,$ where $V_i$ is the eigenspace associated with the eignevalue $\lambda_1.$ This argument is fine and well explained when $\dim(V) = m$ and each eigenspace is $1D.$
The question: Lets say $V_1,$ for concreteness, is 2D (and dim$(V) = m+1$). If all we know is that $V_1$ is both $A$-invariant and $B$-invariant, then it is not necessary that both eigenvectors of $A$ line up with both eigenvectors of $B$ inside $V_1.$ If this is true, how is it possible that a single eigenbasis exists for both?
Thanks in advance
If $V_i$ is the eigenspace of $A$ for the eigenvalue $\lambda_i$, then it's $B$-invariant, as you know. Thus, the matrix of $B$ in a basis of eigenvectors of $A$ writes as $$\begin{bmatrix} B|_{V_1} & & & \\ & B|_{V_2} & & \\ & & \ddots & \\ & & &B|_{V_m} \end{bmatrix}$$ ($B$-invariance means precisely that everything else in this matrix is 0).
Now, the minimal polynomial $\mu_i$ of $B|_{V_i}$ divides the minimal polynomial $\mu_B$ of $B$ since $\mu_B(B|_{V_i})=0$ (because applying $\mu_B$ to $B$ is equivalent to applying it to each block). Hence, $\mu_i$ has simple roots (because it's true for $\mu_B$ since $B$ is diagonalizable) and hence $B_i$ is diagonalizable $\forall i$. This proves that you can find a basis of each of the $V_i$'s containing eigenvectors of $B|_{V_i}$ which are also eigenvectors of $B$ (and of $A$ by def. of $V_i$).
NB: Of course, if you choose randomly a basis of $V_i$, it's not made with eigenvectors of $B$.