Proof of Hermite-Minkowski's Theorem regarding finite number fields for a given discriminant. I want to prove the following theorem by Hermite and Minkowski:

For any given discriminant there are at most finitely many number fields with this discriminant.

A very helpful step is that if $\Delta$ is the discriminant of a number field of degree $n$ (over $\mathbb{Q}$), then $|\Delta| \ge f(n)$ where $f$ is some term in $n$ which goes to infinity as $n$ becomes large.
Hence, the above theorem is reduced to showing that for any given $n$ and $\Delta$, there are only finitely many number fields of degree $n$ with discriminant $\Delta$. This is where I am stuck (I thought about this for a while and have absolutely no idea how to tackle this part of the proof).
Can someone help me out with this one, please? I have the ambition to find the solution on my own, so I'd prefer little hints rather than a full solution.
 A: Note: I tried to formulate a series of hints for you, but with the approach I know and made use of I figured that would simply be more confusing than anything. So here's my proof picking up from where you were unsure how to continue, if you're still interested in the proof. 

As you have determined, for a proof it suffices to show that there are only finitely many solutions to $|\text{disc}(K)| < N$, where we let $K$ denote number fields and $N \in R$. 
Let's consider $K(\sqrt{-1})$. We have two cases, either $K = K(\sqrt{-1})$ (i.e. our discriminant remains unchanged), or $[K(\sqrt{-1}) : K] = 2 $ (i.e. our discriminant changes by bounded amount). We can consequently restrict our attention to $K$ of degree $n$ with $\sqrt{-1} \in K$. Then note that as there are finitely many $d \in \mathbb{Z}, 0 \leq d \leq N$, we can also fix $d$. 
So for our fixed $d$ and $n$, it is enough to prove that there are finitely many $K \ni \sqrt{-1}$ with $|\text{disc}(K)| = d$. Notice that in this case, the Minkowski space is $\mathbb{C}^{\frac{n}{2}}$, and the lattice is $\phi({\mathscr{O}}_{K})$. Now we need to find a centrally symmetric, convex subset such that for any solution $K$, there exists a non-zero point in $B \cap L$, where $B$ is some convex body. Then based on this convex body, we will show that there are only finitely many ways of finding the correct $K$, and then show that in $L \cap B$ we have that it is the image of a unique $K$. 
Let $B = \{(z_{1}, ..., z_{n/2} : |\mathscr{T}_{m}(z_{1})| < c\sqrt{d},|\mathscr{R}_{e}(z_{1})| < 1 |, |z_{i}| < 1, i = 2, ..., \frac{n}{2}\}$. Now pick $L$ sufficiently large so that $\text{vol}(B) \geq 2^{n/2} \cdot \det(L) \geq 2^{n/2}\sqrt{|d|}$. Note that $c$ depends on $n$ and $d$ but not on $K$. 
Now given a solution, we know good bounds on the roots of its minimal polynomial, which gives us good bounds on the coefficients of an integer polynomial. There are only finitely many such minimal polynomials. Therefore, there are only finitely many images of $K$ mapped to $B\cap L$. It remains to prove that one of the $(z_{1}, ..., z_{n/2}) \in B\cap L$ corresponds to a unique $K$. We know that $\mathbb{Q}(z_{1}) \subseteq K$. If $\mathbb{Q}(z_{1}) \neq K$, then there exists $i \neq 1$ such that $\phi_{i}(z_{1}) = z_{i}$. So then $z_{1} = z_{i}$ for some $i$, and we know that $|z_{i}| < 1$. This means that 
$$
\mathbb{N}(z_{1}) = \bigg|\prod_{i=1}^{n/2} z_{i}\bigg| < 1,
$$
which is clearly a contradiction. Therefore $\mathbb{Q}(z_{1}) = K$. Thus there are a finite number of $K = K(\sqrt{-1})$ with $|\text{disc}(K)| = d$ of degree $n$. Hence, we have that there are only finitely many $K$ with $|\text{disc}(K)| < N$, as desired. 
$$ \square $$
