[proofplan]
The proof proceeds via the Dirichlet class number formula, which expresses $h(d)$ in terms of the special value $L(1, \chi_d)$ of the Dirichlet $L$-function attached to the Kronecker symbol $\chi_d = \left( \frac{d}{\cdot} \right)$. For fundamental $d < 0$ the formula reads $h(d) = \frac{w_d \sqrt{|d|}}{2\pi} L(1, \chi_d)$ with $w_d \in \{2, 4, 6\}$ (the number of roots of unity in the order of discriminant $d$), so proving $h(d) \to \infty$ reduces to proving $\sqrt{|d|}\, L(1, \chi_d) \to \infty$ as $d \to -\infty$ through fundamental discriminants. The Heilbronn–Deuring–Landau argument splits into a dichotomy on the zeros of $L(s, \chi_d)$. Case 1: no real zero of $L(s, \chi_d)$ is too close to $s = 1$. Then a standard lower bound $L(1, \chi_d) \gg (\log |d|)^{-1}$ gives $\sqrt{|d|}\, L(1, \chi_d) \to \infty$. Case 2: a real zero $\beta_d \to 1^-$ exists (a Siegel zero). Then Deuring's trick combines $L(s, \chi_{d_0})$ for a fixed auxiliary discriminant $d_0$ with $L(s, \chi_d)$: the product $\zeta(s) L(s, \chi_{d_0}) L(s, \chi_d) L(s, \chi_{d_0} \chi_d)$ has non-negative Dirichlet coefficients, and evaluating near $s = \beta_d$ forces $L(1, \chi_d) \gg |d|^{-\varepsilon}$ for every $\varepsilon > 0$ (with ineffective constant), which still gives $\sqrt{|d|}\, L(1, \chi_d) \to \infty$. In either case $h(d) \to \infty$.
[/proofplan]