To show: If $x\in X_{\lambda }$ and $y\in X_{\gamma }$ then $⟨x,y⟩=0\text{.}$
Assume $\lambda \ne \gamma \text{.}$
Assume $x\in X_{\lambda }$ and $y\in X_{\gamma }\text{.}$
Then $$\begin{array}{rcl}{\displaystyle \lambda ⟨x,y⟩}& =& {\displaystyle ⟨\lambda x,y⟩}\\ & =& {\displaystyle ⟨Tx,y⟩}\\ & =& {\displaystyle ⟨x,T^{*}y⟩}\\ & =& {\displaystyle ⟨x,Ty⟩\text{since}\hspace{0.17em}T\hspace{0.17em}\text{is self adjoint,}}\\ & =& {\displaystyle ⟨x,\gamma y⟩}\\ & =& {\displaystyle \bar{\gamma }⟨x,y⟩}\\ & =& {\displaystyle \gamma ⟨x,y⟩,}\end{array}$$ since $\gamma \in ℝ\text{.}$
So $(\lambda -\gamma )⟨x,y⟩=0\text{.}$
So $\lambda -\gamma =0$ or $⟨x,y⟩=0\text{.}$
Since $\lambda \ne \gamma$ then $⟨x,y⟩=0\text{.}$

$\square$

Assume $\mid m\mid \le M$ (otherwise replace $\mathrm{\Lambda }$ by $-\mathrm{\Lambda }\text{).}$
If $u,v\in H$ then $$\begin{array}{rcl}{\displaystyle 4⟨Tu,v⟩}& =& {\displaystyle ⟨T(u+v),u+v⟩-⟨T(u-v),u-v⟩}\\ & \le & {\displaystyle M({\Vert u+v\Vert }^2+{\Vert u-v\Vert }^2)}\\ & =& {\displaystyle 2M({\Vert u\Vert }^2+{\Vert v\Vert }^2)\text{.}}\end{array}$$ If $Tu\ne 0$ set $$v=\frac{Tu}{\Vert Tu\Vert }·\Vert u\Vert \text{.}$$ Then $$2\Vert u\Vert \Vert Tu\Vert =2⟨Tu,v⟩\le M({\Vert u\Vert }^2+{\Vert v\Vert }^2)=2M{\Vert u\Vert }^2\text{.}$$ So $$\Vert Tu\Vert \le M\Vert u\Vert ,$$ for all $u\in H\text{.}$
So $$\Vert T\Vert \le M\text{.}$$ Assume $u\in H$ and $\Vert u\Vert =1\text{.}$
To show: $\Vert T\Vert \ge \mid ⟨Tu,u⟩\mid \text{.}$ $$\begin{array}{rcl}{\displaystyle \mid ⟨Tu,u⟩\mid }& \le & {\displaystyle \Vert Tu\Vert ·\Vert u\Vert ,\text{by Cauchy-Schwarz}}\\ & \le & {\displaystyle \Vert T\Vert ,}\end{array}$$ since $$\begin{array}{rcl}{\displaystyle \Vert T\Vert }& =& {\displaystyle \text{sup}\left\{\frac{\Vert Tu\Vert }{\Vert u\Vert }\hspace{0.17em}\right|\hspace{0.17em}u\in H\}}\\ & =& {\displaystyle \text{sup}\left\{\Vert Tu\Vert \hspace{0.17em}\right|\hspace{0.17em}\Vert u\Vert =1\}\text{.}}\end{array}$$ So $\Vert T\Vert \ge M\text{.}$
So $\Vert T\Vert =M\text{.}$

$\square$

Proof by contradiction.
Assume $\text{dim}\left(X_{\lambda }\right)$ is infinite dimensional.
Let $e_1,e_2,\dots$ be an orthonormal sequence in $X_{\lambda }\text{.}$
Then $${\Vert e_m-e_n\Vert }^2={\Vert e_m\Vert }^2+{\Vert e_n\Vert }^2=2$$ and $$\Vert T{e}_m-T{e}_n\Vert ={\Vert \lambda e_m-\lambda e_n\Vert }^2={\mid \lambda \mid }^2·2=2{\mid \lambda \mid }^2.$$ So $T{e}_1,T{e}_2,\dots$ does not have a convergent subsequence.

$\square$

Let $X_{\lambda }=\{x\in H\hspace{0.17em}|\hspace{0.17em}Tx=\lambda x\}\text{.}$

(A)	If $\lambda \ne \gamma$ then $X_{\lambda }\perp X_{\gamma }\text{.}$ Proof. Let $x\in X_{\lambda }$ and $y\in X_{\gamma }\text{.}$ Then $$\begin{array}{rcl}{\displaystyle \lambda ⟨x,y⟩}& =& {\displaystyle ⟨\lambda x,y⟩}\\ & =& {\displaystyle ⟨Tx,y⟩}\\ & =& {\displaystyle ⟨x,T^{*}y⟩}\\ & =& {\displaystyle ⟨x,Ty⟩\text{since}\hspace{0.17em}T\hspace{0.17em}\text{is self adjoint}}\\ & =& {\displaystyle ⟨x,\gamma y⟩}\\ & =& {\displaystyle \bar{\gamma }⟨x,y⟩}\\ & =& {\displaystyle \gamma ⟨x,y⟩}\end{array}$$ since eigenvalues of a self adjoint operator are in $ℝ\text{.}$ So $(\lambda ,\gamma )⟨x,y⟩=0\text{.}$ So $\lambda -\gamma =0$ or $⟨x,y⟩=0\text{.}$ So $\lambda =\gamma$ or $⟨x,y⟩=0\text{.}$ $\square$
(A')	If $\lambda \ne 0$ then $\text{dim}\hspace{0.17em}{X}_{\lambda }<\infty \text{.}$
(B)	Choose an orthonormal basis $B_{\lambda }$ of $X_{\lambda }\text{.}$ Let $$B=\underset{\lambda \in \mathrm{\Lambda }}{⨆}{B}_{\lambda }$$ where $\mathrm{\Lambda }=\left\{\text{eigenvalues of}\hspace{0.17em}T\right\}\text{.}$ To show: $H=\overline{\text{span}\hspace{0.17em}B}\text{.}$

$\square$