Vrftsjtryk

I'm facing the following problem:

Let $X$ be a Hilbert space, $A,B$ be subspaces of $X$ such that $X=Aoplus B$ and $Tin B(X)$ such that $Tvert_A=text{Id}_A$, $Tvert_B=-text{Id}_B$. Prove that $T=T^{-1}$ and that $T^*=T$.

The first statement is easy; indeed, if $x=a+b$, then $Tx=a-b$, therefore $TTx=a+b=x$, thus $TT=text{Id}_X$ and we are done.
But I really can't move on with the second statement. I proved that $T^*=T$ if and only if $Abot B$:

Indeed, if $T^*=T$, then $<Tx,y>=<x,T^*y>=<x,Ty>$ for all $x,y$, which yields $<Tx-x,y+Ty>=0$. If $a,bin A,B$ respectively, then for $x=b, y=a$ we have that $<a,b>=0$.
The converse is obvious by taking random $x,y$ and writing them in the form $a+b$ where $ain A, bin B$. But I can't prove that $Abot B$. I tried taking the closure of $A$ and writing $X$ as $X=overline{A}oplusoverline{A}^{bot}$ ,and, since $overline{A}^bot=A^bot$, $X=overline{A}oplus A^{bot}$ but I couldn't move on. Any ideas?

Your problem in understanding here might be that "verticality" of $A$ and $B$ are due to the definition of the direct sum of Hilbert spaces.

For arbitrary Hilbert spaces $(A,langlecdot,cdotrangle_A)$, $(B,langlecdot,cdotrangle_B)$, one defines the direct sum as ordered pairs of the form
$$
Aoplus B=lbrace (a,b),|,ain A,bin Brbrace
$$
which turns into a vector space with componentwise addition and scalar multiplication. Then $Aoplus B$ becomes a Hilbert space under the scalar product
$$
biglangle (a_1,b_1),(a_2,b_2)bigrangle_{Aoplus B}:=langle a_1,a_2rangle_A+langle b_1,b_2rangle_B,.tag{1}
$$
Now it is customary to write these elements of not as ordered pairs $(a,b)$, but as a sum $a+b$ (by identifying $A$ with $Atimeslbrace 0rbracesubset Aotimes B$ and same with $B$). With this identification, every $xin Aoplus B$ can be uniquely expressed as $x=a+b$ for some $ain A$, $bin B$. So in some sense it is by definition that
$$
langle x,Tyrangle=biglangle a_1+b_1,(operatorname{Id}_Aa_2-operatorname{Id}_Bb_2)bigrangle=langle a_1+b_1,a_2-b_2rangleoverset{(1)}=langle a_1,a_2rangle-langle b_1,b_2rangle=ldots= langle Tx,yrangle,,
$$
because $Aoplus B$ by definition (1) / construction / assumption are "orthogonal".

Your arguments are all correct and You have to assume $Aoplus B$ really means the orthogonal sum. If it is not orthogonal by construction as in Frederiks answer and You take the definition $Aoplus B={a+b:ain A,bin B}$ where $Acap B={0}$ then You get simple counterexamples like the finite-dimensional $X=mathbb{R}^2=langlebegin{pmatrix}1\1end{pmatrix}rangleopluslanglebegin{pmatrix}0\1end{pmatrix}rangle$ and $T(begin{pmatrix}1\1end{pmatrix})=begin{pmatrix}1\1end{pmatrix}$ and $T(begin{pmatrix}0\1end{pmatrix})=-begin{pmatrix}0\1end{pmatrix}$. Then $T$ is a bounded operator that fulfills $T_{|A}=Id_A$ for $A=langlebegin{pmatrix}1\1end{pmatrix}rangle$ and $T_{|B}=-Id_B$ for $B=langlebegin{pmatrix}0\1end{pmatrix}rangle$ but is not self-adjoint, what You proved and can check by finding its matrix to an orthogonal basis (e.g. to the standard basis, this yields $begin{pmatrix}1&0\2&-1end{pmatrix}$)