A strict occupied gap defines a unique rank-r spectral projector

By [F1], choose an orthonormal eigenbasis $(u_1,\dots ,u_n)$ of $F(X)$ with the eigenvalues in the given order. The strict gap prevents an eigenspace for one eigenvalue from straddling positions $r$ and $r+1$, so the sum of the first $r$ eigenspaces is the span of exactly the first $r$ basis vectors.

Changing the orthonormal basis within a repeated eigenspace below the gap leaves that entire eigenspace in the sum; changing one within a repeated eigenspace above the gap leaves that entire eigenspace out. Thus the selected subspace does not depend on the eigenbasis chosen in step 1.1. Its orthogonal projector is $P={\sum }_{j=1}^r{u}_j{u}_j^{\ast }$, which satisfies $P=P^{\ast }$ and $P^2=P$ by orthonormality. By [F3], $\mathrm{tr}P={\sum }_{a=1}^n{P}_{aa}={\sum }_{a=1}^n{\sum }_{j=1}^r(u_j{)}_a\overline{(u_j{)}_a}={\sum }_{j=1}^r{\sum }_{a=1}^n\mathrm{\mid }(u_j{)}_a{\mathrm{\mid }}^2={\sum }_{j=1}^{r}1=r$, since each $u_j$ is a unit vector.

The operator in step 2.1 is independent of all allowed basis choices, lies in ${\mathcal{P}}_r$ by [L1], and is exactly the map specified by [F2]. Hence it is the unique rank-$r$ occupied spectral projector.