Spectral theorem for compact self-adjoint operators

First, every eigenvalue of a self-adjoint operator is real: if $Tx=\lambda x$ with $x\ne0$, then $\lambda\|x\|^2=\langle Tx,x\rangle=\langle x,Tx\rangle=\bar\lambda\|x\|^2$ (using self-adjointness on the middle step), so $\lambda=\bar\lambda$. Similarly, eigenvectors for distinct eigenvalues $\lambda\ne\mu$ are orthogonal: $\lambda\langle x,y\rangle=\langle Tx,y\rangle=\langle x,Ty\rangle=\mu\langle x,y\rangle$ forces $\langle x,y\rangle=0$.

Next, compactness guarantees an eigenvalue of maximal absolute value actually exists: the operator norm satisfies $\|T\|=\sup_{\|x\|=1}|\langle Tx,x\rangle|$ for self-adjoint $T$, and compactness lets one extract a convergent subsequence from a maximizing sequence for $|\langle Tx,x\rangle|$, producing a unit vector $e_1$ with $Te_1=\lambda_1 e_1$ where $|\lambda_1|=\|T\|$.

Now induct: having found orthonormal eigenvectors $e_1,\dots,e_{n-1}$ with eigenvalues $\lambda_1,\dots,\lambda_{n-1}$, restrict $T$ to the closed subspace $H_n=\{e_1,\dots,e_{n-1}\}^\perp$. Because $T$ maps $H_n$ into itself (self-adjointness makes the orthogonal complement of an invariant subspace invariant too) and remains compact and self-adjoint there, the same maximal-eigenvalue argument produces the next eigenvector $e_n\in H_n$ with $|\lambda_n|\le|\lambda_{n-1}|$.

Finally, if this process does not terminate, $\lambda_n\to0$: otherwise infinitely many $|\lambda_n|$ would stay above some $\delta>0$, but then $\{Te_n\}=\{\lambda_ne_n\}$ would have no convergent subsequence (since $\|\lambda_ne_n-\lambda_me_n\|^2\ge\delta^2\cdot2$ for $n\ne m$ by orthonormality), contradicting compactness of $T$. One then checks that the closed span of $\{e_n\}$ together with $\ker T$ exhausts $H$, and that $T$ acts as $0$ on $\ker T$, giving the eigen-expansion $Tx=\sum_n\lambda_n\langle x,e_n\rangle e_n$ for all $x\in H$.