As I understand, the idea of Dedekind domains is motivated by the wish to factorize ideals into prime ideals. Dedekind rings are supposed to:
be noetherian, which makes sense because that ensures that the factorization is finite;
have every prime ideal to be a maximal ideal, which makes sense because we want to factorize into prime ideals, so they need to be "very" big.
integral closed.
Can anybody give me an intuitive idea about why we need that property?
On
One of the things that you want in Dedeking rings is that you have nice ideal factorisation properties. In particular, multiplication by a nonzero ideal should be injective.
Suppose $R$ is a commutative ring. Let us look at an element $x/y$ of its field of fractions who happens to be a solution of a monic polynomial equation :
Let $x,y \in R$ and $a_k \in R$ for $k=0 \ldots n-1$ and suppose the equation $x^n = \sum a_k x^k y^{n-k}$ is true.
Now consider $I = (x,y)$, the ideal generated by $x$ and $y$. Then, $I^n = (x^n,x^{n-1}y, \ldots, xy^{n-1},y^n)$. But, because of that equation, the first generator is superfluous, and so $I^n = (x^{n-1}y,\ldots, xy^{n-1},y^n)$. If you look at this ideal carefully, you get $I^n = (y)I^{n-1}$.
Now, if you want your ideals to have nice factorization properties, and if $I$ is nonzero you would want $(y) = I$. Since $I = (x,y)$, this is equivalent to $x \in (y)$, so that there exists a $z \in R$ such that $x=yz$, or also that $x/y \in R$.
Therefore if you want nice ideal factorisations, you need $R$ to be integrally closed.
(checking that a few ideals factor as they should is also what I do when I want to check if a ring is integrally closed)
Also, wikipedia lists various alternative definitions, some of which place the focus on ideal factorization :
"every proper ideal factors into primes"
"every nonzero fractional ideal is invertible"
Those may be more to your liking.
Integral closure is needed to get factorization of ideals into products of prime ideals.
Consider the ring $R=k[t^2,t^3]$ where $k$ is some field and $t$ is an indeterminate. The ideal $P=(t^2,t^3)$ is prime, it consists of those polynomials in $R$ that vanish at $t=0$. It is also a maximal ideal because $R/P\simeq k$ is a field.
We see that the ideal $I=(t^3,t^4)$ is contained in $P$. Yet it is not any power of $P$ either ($P^2=(t^4,t^5)$ is already a proper subset of $I$). It follows easily that $I$ is not a product of prime ideals of $R$.
Of course, moving to the integral closure $\overline{R}=k[t]$ in the field of fractions of $R$ fixes this problem. Then the relevant prime ideal is $\mathfrak{p}=(t)$ and both $(t^2,t^3)=(t^2)$ and $(t^3,t^4)=(t^3)$ become powers of $\mathfrak{p}$.
Similar things happen with non-maximal orders of number fields.
(Extras) I used the ring $R$ as an example because it lets me point out the following connection. Namely, the ring $R$ is isomorphic to the coordinate ring $\Gamma(C)$ of the plane curve $C:y^2=x^3$, $$\Gamma(C)=k[x,y]/(y^2-x^3),$$ the isomorphism given by $y\mapsto t^3$, $x\mapsto t^2$. Here the integral non-closure shows up geometrically in the cusp at the origin $O=(0,0)$. By calculating the partial derivatives of $y^2-x^3$ at the origin you see that the origin is not a smooth point (= a point where implicit function theorem can be applied). It is a fact that a plane curve defined by a polynomial equation has no singular points if and only if its coordinate ring is integrally closed (when it becomes a Dedekind domain).