= Solution
<Minkowski convex body theorem> states that a centrally symmetric <convex> measurable subset of $\mathbb R^n$ with <volume> greater than $2^n$ times the <covolume> of a full <Euclidean lattice> contains a nonzero lattice point. The equivalent closed-boundary form follows by compactness for compact <convex> bodies.
The set $S_t$ is centrally symmetric. It is <convex> because for $0\le\lambda\le1$, the <triangle inequality> gives $\|\lambda u+(1-\lambda)v\|_1\le\lambda\|u\|_1+(1-\lambda)\|v\|_1\le t$. It is a diamond of area $2t^2$. For the embedded nonzero <integral ideal> $I$, the given <covolume> is $2N(I)\sqrt d$. Hence <Minkowski convex body theorem> applies when
$$
2t^2>8N(I)\sqrt d,\qquad t^2>4N(I)\sqrt d.
$$
For the resulting nonzero $\alpha\in I$, write $x=\sigma_1(\alpha)$, $y=\sigma_2(\alpha)$. Then
$$
|\operatorname{Norm}_{K/\mathbb Q}(\alpha)|=|xy|
\le\frac{(|x|+|y|)^2}{4}\le\frac{t^2}{4}.
$$
Decrease $t$ to $2\sqrt{N(I)\sqrt d}$. A fixed slightly larger diamond contains only finitely many lattice points, so one nonzero $\alpha$ recurs along a sequence of such $t$ values. Passing to the limit proves
$$
\boxed{|\operatorname{Norm}_{K/\mathbb Q}(\alpha)|\le N(I)\sqrt d.}
$$
Given a target <ideal class> $C$, choose an <integral ideal> $I$ in $C^{-1}$. Since $\alpha\in I$, the <ideal> $J=(\alpha)I^{-1}$ is integral, belongs to $C$, and has <ideal norm>
$$
N(J)=\frac{|\operatorname{Norm}(\alpha)|}{N(I)}\le\sqrt d.
$$
For $K=\mathbb Q(\sqrt{14})$, every <ideal class> thus has an integral representative of norm at most $\sqrt{14}<4$. The only possibilities are norms one, two or three. The prime two ramifies, giving the unique <ideal> of norm two $P_2=(2,\sqrt{14})$. The element $4+\sqrt{14}$ has <field norm> $16-14=2$, so $(4+\sqrt{14})=P_2$, making it principal. The prime three is inert because $X^2-14\equiv X^2+1\pmod3$ is an <irreducible polynomial>; its <prime ideal> has norm nine, so there is no <ideal> of norm three. Every possible small representative is principal, proving <class number one for Q square root of fourteen>. Therefore
$$
\boxed{h(\mathbb Q(\sqrt{14}))=1.}
$$
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