Skip past proof (i), I only included it because the proof following refers back to it. Below, I'm stuck on proof (ii), and the two (one,two) similar posts I found used Rodrigues' formula and a different formula. The red one below appears to come from $P_n=xP_{n-1}$ (I'm guessing so look at the formula and don't be mislead). Where did this $P_n$ come from. Likewise, where did the $P_{n-1}$ at the red 2 come from. Finally why are the right hand sides of the 2 equations before red 1 and red 2 equal?
We can try to add some details to the proof, so maybe you can understand better. From the Bonnet's recurrence relarions $$\left(n+1\right)P_{n+1}+nP_{n-1}=\left(2n+1\right)xP_{n}\tag{1} $$ $$nP_{n}+\left(n-1\right)P_{n-2}=\left(2n-1\right)xP_{n-1}\tag{2} $$ we observe that if we multiply $(1)$ by $P_{n-1} $ and $(2)$ by $P_{n} $ we have $$\left(n+1\right)P_{n+1}P_{n-1}+nP_{n-1}^{2}=\left(2n+1\right)xP_{n}P_{n-1} $$ $$nP_{n}^{2}+\left(n-1\right)P_{n-2}P_{n}=\left(2n-1\right)xP_{n-1}P_{n} $$ hence we have $$\left(n+1\right)\int_{-1}^{1}P_{n+1}\left(x\right)P_{n-1}\left(x\right)dx+n\int_{-1}^{1}P_{n-1}^{2}\left(x\right)dx=\left(2n+1\right)\int_{-1}^{1}xP_{n}\left(x\right)P_{n-1}\left(x\right)dx $$ $$n\int_{-1}^{1}P_{n}^{2}\left(x\right)dx+\left(n-1\right)\int_{-1}^{1}P_{n-2}\left(x\right)P_{n}\left(x\right)dx=\left(2n-1\right)\int_{-1}^{1}xP_{n}\left(x\right)P_{n-1}\left(x\right)dx $$ but from $(i)$ we know that $$\int_{-1}^{1}P_{n+1}\left(x\right)P_{n-1}\left(x\right)dx=\int_{-1}^{1}P_{n-2}\left(x\right)P_{n}\left(x\right)dx=0 $$ so $$n\int_{-1}^{1}P_{n-1}^{2}\left(x\right)dx=\left(2n+1\right)\int_{-1}^{1}xP_{n}\left(x\right)P_{n-1}\left(x\right)dx\tag{3} $$ $$n\int_{-1}^{1}P_{n}^{2}\left(x\right)dx=\left(2n-1\right)\int_{-1}^{1}xP_{n}\left(x\right)P_{n-1}\left(x\right)dx\tag{4} $$ which is equivalent to $$\frac{n}{2n+1}\int_{-1}^{1}P_{n-1}^{2}\left(x\right)dx=\int_{-1}^{1}xP_{n}\left(x\right)P_{n-1}\left(x\right)\tag{5} $$ $$\frac{n}{2n-1}\int_{-1}^{1}P_{n}^{2}\left(x\right)dx=\int_{-1}^{1}xP_{n}\left(x\right)P_{n-1}\left(x\right)dx\tag{6} $$ and now we can observe that the RHS of $(5)$ and $(6)$ are equal, so $$\frac{n}{2n+1}\int_{-1}^{1}P_{n-1}^{2}\left(x\right)dx=\frac{n}{2n-1}\int_{-1}^{1}P_{n}^{2}\left(x\right)dx $$ $$\Leftrightarrow\int_{-1}^{1}P_{n}^{2}\left(x\right)dx=\frac{2n-1}{2n+1}\int_{-1}^{1}P_{n-1}^{2}\left(x\right)dx $$ now we can conclude the proof by induction. Assume that $$\int_{-1}^{1}P_{n-1}^{2}\left(x\right)dx=\frac{2}{2n-1} $$ then $$\int_{-1}^{1}P_{n}^{2}\left(x\right)dx=\frac{2n-1}{2n+1}\frac{1}{2n-1}=\frac{2}{2n+1} $$ and $$\int_{-1}^{1}P_{1}^{2}\left(x\right)dx=\frac{2}{3}.$$