Step 1: Existence of Eigenvalues

Define $\lambda_1 = \sup_{\|x\|=1} |\langle Tx, x \rangle|$. Since $T$ is self-adjoint, $\langle Tx, x \rangle \in \mathbb{R}$ for all $x$.

Choose a sequence $(x_n)$ with $\|x_n\| = 1$ and $|\langle Tx_n, x_n \rangle| \to \lambda_1$. The sequence $(Tx_n)$ is bounded, so by compactness of $T$, there exists a subsequence (still denoted $x_n$) such that $Tx_n \to y$ for some $y \in H$.

By the parallelogram law and properties of the supremum, we can show that $(x_n)$ is Cauchy, hence converges to some $e_1$ with $\|e_1\| = 1$. By continuity, $Te_1 = y$.

From $\langle Tx_n, x_n \rangle \to \pm \lambda_1$ and $Tx_n \to Te_1$, we get $\langle Te_1, e_1 \rangle = \pm \lambda_1$. Setting $\lambda_1$ to be this value (with sign), we have $Te_1 = \lambda_1 e_1$.

Step 2: Orthogonal Decomposition

Let $H_1 = \text{span}\{e_1\}^\perp$ be the orthogonal complement. For any $x \in H_1$, we have $\langle Tx, e_1 \rangle = \langle x, Te_1 \rangle = \lambda_1 \langle x, e_1 \rangle = 0$

Thus $T$ maps $H_1$ to itself. The restriction $T|_{H_1}$ is also compact and self-adjoint.

Step 3: Iteration

Repeat Step 1 on $H_1$ to find $\lambda_2$ and $e_2$, then on $H_2 = \text{span}\{e_1, e_2\}^\perp$, and so on. This produces an orthonormal sequence $\{e_n\}$ of eigenvectors with eigenvalues $\{\lambda_n\}$ satisfying $|\lambda_1| \geq |\lambda_2| \geq \cdots$.

Step 4: Convergence of Eigenvalues

If $\lambda_n \not\to 0$, then for some $\varepsilon > 0$, infinitely many $|\lambda_n| \geq \varepsilon$. The sequence $(e_n)$ satisfies $\|e_n\| = 1$ but $\|Te_n\| = |\lambda_n| \geq \varepsilon$. Since $e_n \perp e_m$ for $n \neq m$, we have $\|Te_n - Te_m\|^2 = \|Te_n\|^2 + \|Te_m\|^2 \geq 2\varepsilon^2$. Thus $(Te_n)$ has no convergent subsequence, contradicting compactness.

Step 5: Completeness

Let $M = \overline{\text{span}\{e_1, e_2, \ldots\}}$. For $x \in M^\perp$, we have $\langle x, e_n \rangle = 0$ for all $n$, so $\langle Tx, e_n \rangle = \lambda_n \langle x, e_n \rangle = 0$. Thus $Tx \in M^\perp$.

If $M \neq H$, then $T|_{M^\perp}$ is a nonzero compact self-adjoint operator, which must have a nonzero eigenvalue $\lambda$ with eigenvalue $e$. But this contradicts the construction. Therefore $M = H$.