Let $[X]$ and $[X_{0}]$ denote the span of $X$ and $X_{0}$, respectively. Further, let $P_{[X_{0}]}$ denote the orthogonal projection onto the subspace $[X_{0}]$, and $P_{[X]}$ denote the orthogonal projection onto the subspace $[X]$. Further assume that $[X_{0}]\subseteq [X]$.
Why does the following equality hold $$ \lvert \lvert Y-P_{[X]}Y\rvert \rvert ^{2}-\lvert \lvert Y-P_{[X_{0}]}Y\rvert \rvert ^{2}=\lvert \lvert P_{[X_{0}]}Y-P_{[X]}Y\rvert \rvert ^{2}\; \; ?$$
My attempt:
$$\lvert \lvert P_{[X_{0}]}Y-P_{[X]}Y\rvert \rvert ^{2}=\lvert \lvert Y-P_{[X]}Y-(Y-P_{[X_{0}]}Y)\rvert \rvert ^{2}\\=\lvert \lvert Y-P_{[X]}Y\rvert \rvert ^{2}+\lvert \lvert Y-P_{[X_{0}]}Y\rvert \rvert ^{2}-2\langle Y-P_{[X]}Y ,Y-P_{[X_{0}]}Y\rangle$$
So I suppose we have to show that $\langle Y-P_{[X]}Y ,Y-P_{[X_{0}]}Y\rangle= \langle Y-P_{[X_{0}]}Y ,Y-P_{[X_{0}]}Y\rangle= \lvert \lvert Y-P_{[X_{0}]}Y\rvert \rvert ^{2}$, but I am unsure whether this has to hold?
I will use $P_0$ and $P_1$ for the projections onto $[X_0]$ and $[X]$ respectively.
The condition you are given, $[X_0]\subset[X_1]$ is that $P_0P_1=P_0$. Then $$ (I-P_0)(I-P_1)=I-P_0-P_1+P_0P_1=I-P_0. $$ This is exactly what you need: \begin{align} \langle y-P_0y,y-P_1y\rangle&=\langle (I-P_0)y,(I-P_1)y\rangle =\langle (I-P_1)^*(I-P_0)y,y\rangle\\[0.3cm] &=\langle (I-P_1)(I-P_0)y,y\rangle =\langle (I-P_0)^2y,y\rangle\\[0.3cm] &=\langle (I-P_0)y,(I-P_0)y\rangle =\|y-P_0y\|^2. \end{align}