It is not very clear to me what you mean by "intersection of Hilbert Class Fields [...] is discussed". The theory of Complex Multiplication (see, for instance, Serre's short note in Cassels and Frohlich's *Algebraic Number Theory*, or Silverman's *Advanced Topics in the Arithmetic of Elliptic Curves*, or directly the bible from Shimura, *Introduction to the Arithmetic Theory of Automorphic Functions*) tells you that there is an explicit way, given an imaginary quadratic field $K=\mathbb{Q}(\sqrt{d})$ to construct not only its Hilbert class field, but all ray class fields of different conductors. For instance, for the Hilbert class field, you can first create the elliptic curve $\mathbb{C}/\mathcal{O}_K$: it is an elliptic curve with complex multiplication by $\mathcal{O}_K$. Then, the theory will tell you that the smallest extension of $\mathbb{Q}$ containing $\sqrt{d}$ over which this curve is defined is the Hilbert Class Field of $K$. Just to be convinced that what I say is believable (beside being true, for which you might look at the references), observe that there is an elliptic curve with CM by $\mathbb{Z}[i]$ which admits a Weierstrass equation
$$
y^2=x^3+x\;.
$$
This is defined over $\mathbb{Q}$, so the smallest field of definition which contains $\sqrt{-1}$ is $\mathbb{Q}(i)$, which is indeed its own Hilbert class field.

In more concrete terms, if you are given a squarefree $d<0$ then the Hilbert class field of $K=\mathbb{Q}(\sqrt{d})$ is$K(j(\sqrt{d})$, where $j$ is the modular function
$$
j(q)=\frac{1}{q}+744+196884q+\dots
$$
whose definition you'll find in the above references, or at
http://en.wikipedia.org/wiki/J-invariant . All the fun is in proving that $j(\sqrt{d})$ is actually an algebraic number (it is even an algebraic integer!); then, of course, one proves the abelian+unramified property of the (now, finite!) extension $K(j(\sqrt{d}))/K$. Quiteremarkably, it is an algebraic number ''because" its conjugates are precisely the $j$-invariants of the elliptic curves $\mathbb{C}/\mathfrak{a}_i$ for $\mathfrak{a}_i$ running through a set of integral representatives of the ideal class. This, already, shows that the degree of the extension coincides with the class number (and rapidly leads to it being at least Galois).

In your case, life is easier: you only want to be sure that if $d\equiv 1\pmod{4}$ is negative and squarefree, then $K(i)/K$ is unramified (being abelian, this would force it to lie inside the Hilbert class field). For this, it is enough to observe that $K(i)/\mathbb{Q}$ is a biquadratic extension with Galois group $(\mathbb{Z}/2)^2$ and has therefore three quadratic subfields: $\mathbb{Q}(i),K$ and $F=\mathbb{Q}(\sqrt{-d})$. Now pick a prime $\ell$ dividing $d$: it is necessarily odd. Its ramification degree is $2$ both in $K$ and in $F$, while it is unramified in $\mathbb{Q}(i)$. Therefore it needs be unramified in $K(i)/K$, since ramification degrees are multiplicative in towers of extensions. Similarly, $2$ is unramified in $F/\mathbb{Q}$ and cannot ramify in $K(i)/K$. For what concerns the infinite primes, observe that $K$ is already totally complex, so no ramification can occur. Done!