How to prove that the set of real $ n \times n $ PSD matrices of rank $\leq r < n $ and unit trace, is not contractible? Extended Version of the Question:
How to prove that the set of real $n \times n$, symmetric positive semidefinite (PSD) matrices of rank $\leq r$ ($ 1 \lt r \lt n-1 $) and unit trace, is not contractible?
Question Preliminaries:
Let $\rho^r_n$ denote the set of real $n \times n$, symmetric positive semidefinite (PSD) matrices that are (a) rank $\leq r$ and (b) unit trace, i.e.,
$$
\rho^r_n := \left\{ X \in \mathcal{S}^{+}_n \; \cap \; \mathcal{S}^{\mathcal{T}1}_n \; | \; rank(X) \leq r \right\}
$$
defined for any $1 \leq r \leq n $.
Where $ \mathcal{S}^{+}_n $ is the set of real $n \times n$, symmetric PSD matrices:
$$
\mathcal{S}^{+}_n := \left\{ X \in \mathcal{S} \; | \; y^T X y \geq 0 \; \forall \; y \in \mathbb{R}^n \setminus \{0\} \right\}
$$
Similarly, denote the set of symmetric positive definite matrices, $ \mathcal{S}^{++}_n $ as:
$$
\mathcal{S}^{++}_n := \left\{ X \in \mathcal{S} \; | \; y^T X y > 0 \; \forall \; y \in \mathbb{R}^n \setminus \{0\} \right\}
$$
$ \mathcal{S}^{\mathcal{T}1}_n $ is the set of real $n \times n$ symmetric matrices with unit trace:
$$
\mathcal{S}^{\mathcal{T}1}_n := \left\{ X \in \mathcal{S}_n \; | \; trace(X)=1 \right\}
$$
And $ \mathcal{S} $ is the set of real $n \times n$ symmetric matrices:
$$
\mathcal{S}_n := \left\{ X \in \mathbb{R}^{n \times n} \; | \; X=X^T \right\}
$$
Some Observations:
(I)  Note that by the definitions above the set $ \rho^n_n $ is the intersection of the real PSD cone and the affine space of real symmetric unit trace matrices:
$$
\rho^n_n = \mathcal{S}^{+}_n \; \cap \; \mathcal{S}^{\mathcal{T}1}_n 
$$
The sets $ \mathcal{S}^{+}_n $ and $ \mathcal{S}^{\mathcal{T}1}_n $ are both convex and so is their intersection, which implies that $ \rho^n_n $ is convex (and bounded).  Since every convex bounded set is contractible (i.e., it is homotopy equivalent to a one-point space) then so is $ \rho^n_n $.

(II)  The set $ \rho^n_n $ has dimension $\frac{n^2 + n}{2} - 1$ (the set $ \mathcal{S}^{+}_n $ has the same dimension of the ambient space $ \mathcal{S}_n $ of $ \frac{n^2 + n}{2} $ and the trace constraint reduces the dimension by $ 1 $).
Since $ \rho^n_n $ is a closed and bouned convex set of dimension $\frac{n^2 + n}{2} - 1$ it is homeomorphic to the closed $\left( \frac{n^2 + n}{2} - 1 \right) $-dimensional ball,  $\overline{\mathbb{B}}^{\frac{n^2 + n}{2} - 1}$.
Let $ \rho^{++}_n $ denote the set of real $ n \times n $ symmetric positive definite matrices:
$$
\rho^{++}_n = \mathcal{S}^{++} \; \cap \; \mathcal{S}_{\mathcal{T}1} 
$$
It is straightforward to see that $ \mathcal{S}^{++} $ is open which implies that $ \rho^{++}_n $ is also open.  It is clear that the closure of $ \rho^{++}_n $ is $ \rho^{n-1}_n $.
The boundary of $\overline{\mathbb{B}}^{\frac{n^2 + n}{2} - 1}$ is the unit sphere of dimension $\frac{n^2 + n}{2} - 1$, $\mathbb{S}^{\frac{n^2 + n}{2} - 1}$.
I would argue that $ \rho^{n-1}_n $ is equivalent to $\mathbb{S}^{\frac{n^2 + n}{2} - 1}$ since it is the closure of $ \rho^{++}_n $, which is equivalent to the open unit ball $\frac{n^2 + n}{2} - 1$.
$\mathbb{S}^{\frac{n^2 + n}{2} - 1}$ is not contractible for $ n \gt 1 $ which implies that neither is $ \rho^{n-1}_n $.

(III)  It is straightforward to shown that $ \rho^1_n $ is topologically equivalent to $ \mathbb{RP}^{n-1} $, the real projective space of dimension $n-1$:
$$
\rho^1_n = \{ y y^T \; | \; y \in \mathbb{R}^n, \; \| y \|_2 = 1 \}
$$
and note that $ y y^T = (-y) (-y)^T $ provides the necessary equivalence relation on $ \mathbb{R}^n $:
$$
x \sim y \Leftrightarrow x=-y
$$
and $ \mathbb{RP}^{n-1} = \left\{ \mathbb{R}^n \setminus {0} \right\} / \sim $.
Since $ \mathbb{RP}^{n-1} $ is not homotopy equivalent to a point it is clear that $ \rho^1_n $ is not contractible.

Questions:
(1) How can one show that the set $ \rho^r_n $ is either contractible or not for $ 1 \lt r \lt n-1 $?
(2) I have a hunch that these sets are not contractible so how would one go about showing that?
I realize the most straightforward way would be to show that these sets are not homotopy equivalent to a point but not being well-versed in topology I am not even sure how to go about this.
(3) Any ideas about extending this to the complex case?
Please help!
 A: Disclaimer: I'm not an expert when it comes to topology, so be careful with this answer.
Let $A_r$ be one such matrix (real or complex, positive semidefinite, with unit trace). Then it has the eigenvalue decomposition:
$$A_r = U \Lambda_r U^*,$$
where $\Lambda_r = \operatorname(\lambda_1,\dots,\lambda_r,0,\dots,0)$, $\lambda_k \ge 0$, $\sum_k \lambda_k = 1$, and $U$ is unitary (orthogonal in the real case). There is an obvious continuous map from $\Lambda_r$ to
$$\Lambda_{r-1} = \operatorname{diag}(\lambda_1+\lambda_r,\lambda_2,\dots,\lambda_{r-1},0,\dots,0),$$
defining a unit-trace PSD matrix $A_{r-1} := U \Lambda_{r-1} U^*$ of rank at most $r-1$.
Continue like that until you get $A_1 := U \operatorname{diag}(1,0,\dots,0) U^*$. It remains to show that there is a continuous map between any two such matrices, i.e., that all unit-trace rank-one PSD matrices are homotopic.
For that to be true, you need the set of unitary matrices to be path-connected. This is true for unitary matrices, but not for orthogonal matrices (which also gives you a way to find a counter example to show that this is really not a contractible set).
Note: I noticed that you disallow the rank to drop to $1$, but that can be easily avoided by noticing that all unit-trace PSD matrices of rank $2$ have the form $U \operatorname{diag}(\lambda_1,1-\lambda_1,0,\dots,0) U^*$ for some unitaty matrix $U$, and positive number $\lambda_1 < 1$, making my answer a bit more complex but essentially the same.
Note 2: A real case might still be salvageable, at least in the case of odd dimension, by multiplying the orthogonal factor with $-1$.
