Related this this question.
I'm not able to work out why the manifold
$$ M = \left\{A \in \mathbb{R}^{3 \times 3} : A^2 - A = 0 \right\} $$
The only observation I was able to make is that the eigenvalues of the polynomial
$$ p(A) = A^2 - A $$
Are constructed by evaluating the polynomial
$$ p(\lambda) = \lambda^2-\lambda $$
at the eigenvalues of $A$. But I'm not really sure how I should use this and even if I can. The other observation is that the roots of $p(\lambda)$ are $\left\{0, 1 \right\}$. But even here I'm not really able to use this.
Are you asking to show that $M$ is disconnected? We have a continuous map $\text{det}: \mathbb{R}^{3 \times 3} \to \mathbb{R}$, and the image of $M$ lands on the discrete set $\{0, 1\}$, so there are at least two connected components.
In fact there are four connected components, as explained in the comments on the other question.
EDIT: More detail on why there are exactly four: from looking the trace as a map to $\mathbb{R}$ we see there are at least four connected components. In fact these are it:
any matrix in $M$ is conjugate to a matrix whose diagonal entries are all zeroes and ones, and we can assume the entries are in (weakly) increasing) order. Let's write $D_{0, 0, 0}$, $D_{0, 0, 1}$, $D_{0, 1, 1}$, and $D_{1, 1, 1}$ for these diagonal matrices, and $C_{*, *, *}$ for their respective conjugacy classes.
For each of these conjugacy classes, we have a surjection
$$ \text{GL}^+_3(\mathbb{R}) \to C_{*, *, *} $$ given by $\gamma \mapsto \gamma D_{*, *, *} \gamma^{-1}.$
This proves the conjugacy classes are connected, since the domain is.