 Solution

Self-adjointness means $⟨Tv,w⟩=⟨v,Tw⟩$ for all $v,w\in V$, or $T=T^{\ast }$. Take the inner product linear in its first variable. If $Tv=\lambda v$ with $v\ne 0$, thenTherefore every eigenvalue is real.

If $U$ is invariant and $w\in U^{\perp }$, then for every $u\in U$, $Tu\in U$ and $⟨Tw,u⟩=⟨w,Tu⟩=0$. Hence $Tw\in U^{\perp }$, proving invariance of the orthogonal complement.

Induct on $d=\dim V$. The zero-dimensional case is empty. For $d>0$, choose an eigenvector and normalize it to a unit vector $e$. Its span is invariant, so its orthogonal complement is invariant as just proved. The restriction of $T$ to that complement is again self-adjoint. Its dimension is $d-1$, so induction supplies an orthonormal basis of eigenvectors there. Adjoining $e$ gives an orthonormal eigenbasis of $V$. This proves the finite-dimensional spectral theorem without assuming it in the argument.