[Now fixed and completed :-)]

I think the following roundabout argument works when $d\equiv 5\pmod{10}$ is square-free.

Assume $d>0$. My solution needs some facts from class field theory, because the basic idea is to show that $K=\Bbb{Q}(\sqrt{d},\sqrt5)$ is a quadratic, unramified, totally real extension of $L=\Bbb{Q}(\sqrt d)$. As $K/L$ is abelian, class field theory then implies that $K$ is contained in the Hilbert class field $H$ of $L$. But it is known that $[H:L]=h$ is the class number of $L$, so we get as a by-product that the class number of $L$ is even. The immediate corollary is that $\mathcal{O}_L$ cannot be a UFD.

Write $d=5a$. The claim follows from a calculation of discriminants, but the details of the calculation differ according to the residue class of $a$ modulo $4$.

**Case I.** I assume first that $a\equiv-1\pmod4$, or that $d\equiv-5\pmod{20}$. It is well known that $\Delta_{L/\Bbb{Q}}=4d=20a$, because $d\not\equiv1\pmod4$. On the other hand $K$ contains $\Bbb{Q}(\sqrt{5})$ and $\Bbb{Q}(\sqrt{a})$ as subfields. So if we denote $\phi=(1+\sqrt{5})/2$ we know that the $\Bbb{Z}$-span of $S=\{1,\phi,\sqrt a,\phi\sqrt a\}=\{s_1,s_2,s_3,s_4\}$ is contained in $\mathcal{O}_K$. Therefore the determinant of the matrix $A=(tr(s_is_j)_{1\le i,j\le 4})$ has the discriminant $D=\Delta_{K/\Bbb{Q}}$ as a factor. But, unless I made a mistake
$$
A=\left(\begin{array}{cccc}
4&2&0&0\\
2&6&0&0\\
0&0&4a&2a\\
0&0&2a&6a\end{array}\right),
$$
and this has determinant $400a^2=(4d)^2$. Because the relative discriminant $\Delta_{K/L}$ (necessarily viewed as an ideal) satisfies the relation
$$
D=N_{L/\Bbb{Q}}(\Delta_{K/L})(\Delta_{L/\Bbb{Q}})^2=16d^2 N_{L/\Bbb{Q}}(\Delta_{K/L})\mid\det A= 16d^2,
$$
we can conclude that $S$ must be an integral basis and that the relative discriminant is $1$. The claim follows.

**Case II.** Assume next that $a\equiv 1\pmod 4$, $a\neq1$. This time $d\equiv1\pmod4$, and we know that $d(L/\Bbb{Q})=d=5a$. Again $K$ has the subfields $\Bbb{Q}(\sqrt 5)$ and $\Bbb{Q}(\sqrt a)$, but this time both of them have "smaller" algebraic integers. If we abbreviate $\theta=(1+\sqrt a)/2$, we know that $S=\{s_1,s_2,s_3,s_4\}=\{1,\phi,\theta,\phi\theta\}$ is a set of algebraic integers of $K$, linearly independent over $\Bbb{Z}$. This time the matrix
$$
A=(tr^K_{\Bbb{Q}}(s_is_j))_{1\le i,j\le4}
=\left(\begin{array}{cccc}
4&2&2&1\\
2&6&1&3\\
2&1&1+a&\frac{1+a}2\\
1&3&\frac{1+a}2&\frac{3(1+a)}2\end{array}\right).
$$
Again we are lucky, and $\det A=25a^2=d(L/\Bbb{Q})^2.$ The claim follows as in Case I.

A similar argument shows that $\Bbb{Q}(\sqrt{d},i)/\Bbb{Q}(\sqrt d)$ is unramified at all finite primes. Initially I mistakenly thought that this again implies that the Hilbert class field of $\Bbb{Q}(\sqrt d)$ contains $\Bbb{Q}(\sqrt{d},i)$. However, the infinite primes fail here unless we also have $d<0$, so this does not follow (unless $d<0$ which was excluded in the question).

But, the claim also holds when $d<0$. The above calculations of discriminants are immune to the sign of $d$. The only change is in the class field theoretical parts. Because $\Bbb{Q}(\sqrt d)$ is now totally imaginary, the question about the ramification of the infinite prime does not arise at all! The claim holds for all $d\equiv5\pmod {10}, d\neq 5.$