Let $X,Y$ be two linear subspace of the same dimension of a finite linear space. Denote by $\|\cdot\|_2$ the induced matrix norm. The question asks me to prove the following, as written in title. $$\|P_X-P_Y\|_2=\|(1-P_Y)P_X\|_2$$
where $P_{X}$ refers to the projection operator which satisfy $(P_X)^*=P_X=(P_X)^2$ and $\text{image}(P_X)=X$.
As far as I consider, the projection matrices are Hermitian, thus it can be written in diagonal form $P_X=UAU^*,\,P_Y=VBV^*$. Furthermore, since they're of the same dimension, we can assume $A=B=L$ is the diagonal matrix whose first $k$ diagonal element is $1$ and $0$ otherwise.
Notice $$P_X-P_Y=(1-P_Y)P_X-P_Y(1-P_X):=A+B$$ where the two matrices are orthogonal to each other, $A^*B=B^*A=0$.
$$\|P_X-P_Y\|_2=\|(1-P_Y)P_X\|_2\lor\|P_Y(1-P_X)\|_2$$
It's now equivalent to prove $$\|(1-P_Y)P_X\|_2\geq \|P_Y(1-P_X)\|_2$$
With the decomposition of Hermitians, it's equivalent to verify $$\|LW(I-L)\|_2\geq \|LW^*(I-L)\|_2$$
where $W=U^*V$.
It seems in fact the inequality should hold as an equality. But I have no idea how to proceed.
Note $$A:=LW(I-L),\, B:=(I-L)WL$$ can be considered as submatrices of the unitary matrix $W$.
Denote $C=LWL$. We have $$AA^*=L-CC^*,\, B^*B=L-C^*C$$ where $L$ is the identity map of subspace spanned by first $k$ columns, which is sufficient to deduce $AA^*,B^*B$ share the eigenvalues, to be explicit, $\sqrt{1-\sigma^2(C)}$, and $A,\,B$ share the singular values (other than mutilpicity of $0$ to be rigorous).