The idempotents 0 and I are isolated points

Idea. On $\mathcal E$ the continuous function $\operatorname{tr}$ takes only integer values, hence is locally constant. The values $0$ and $n$ are attained at a single point each.

Step 1: $\operatorname{tr}P=\operatorname{rank}P$ for $P\in\mathcal E$. From $P(P-I)=0$, the minimal polynomial of $P$ divides $t(t-1)$, which has distinct roots, so $P$ is diagonalisable with eigenvalues in $\{0,1\}$. If $P=X\,\mathrm{diag}(I_r,0)X^{-1}$ then $\operatorname{rank}P=r$ and $\operatorname{tr}P=r$.

Step 2: local constancy. The map $\operatorname{tr}:M_n\to\mathbb{R}$ is continuous (linear), and it takes values in $\{0,1,\dots,n\}$ on $\mathcal E$. Thus
$$\mathcal E\cap\{\operatorname{tr}<\tfrac12\}=\{P\in\mathcal E:\operatorname{tr}P=0\}=\{P\in\mathcal E:\operatorname{rank}P=0\}=\{0\},$$
and this set is open in $\mathcal E$ (it is the intersection of $\mathcal E$ with an open subset of $M_n$). So $0$ is isolated. Likewise
$$\mathcal E\cap\{\operatorname{tr}>n-\tfrac12\}=\{P\in\mathcal E:\operatorname{rank}P=n\}=\{I\},$$
since an invertible idempotent satisfies $P=P^{-1}P^2=I$. So $I$ is isolated.

Alternative quantitative proof. If $P\in\mathcal E$ and $P\ne0$, then $\|P\|=\|P^2\|\le\|P\|^2$ and $\|P\|>0$, so $\|P\|\ge1$ in the operator norm. Hence every idempotent other than $0$ is at distance $\ge1$ from $0$. Applying this to the idempotent $I-P$ (indeed $(I-P)^2=I-2P+P=I-P$) shows $\|I-P\|\ge1$ for $P\ne I$. So the balls of radius $1$ around $0$ and around $I$ contain no other idempotent.

Remark. The same reasoning shows that $\mathcal E=\bigsqcup_{r=0}^n\mathcal E_r$ with $\mathcal E_r=\{\operatorname{rank}=r\}=\{\operatorname{tr}=r\}$, each $\mathcal E_r$ being both open and closed in $\mathcal E$. For $0<r<n$ each $\mathcal E_r$ is connected: it is the orbit of $\mathrm{diag}(I_r,0)$ under conjugation by $GL_n(\mathbb{R})$, and the stabiliser contains $\mathrm{diag}(-1,1,\dots,1)$ of determinant $-1$, so the connected group $GL_n^+$ already reaches the whole orbit. So $\mathcal E$ has exactly $n+1$ components.

Lesson. A continuous function with values in a discrete set is locally constant; use a continuous integer invariant (trace/rank) to detect isolated points.