The formal definition of an interval 
I is A real interval iff ∀ x,y ∈ I the segment  [x,y] ⊂ I

I can't understand why an interval is defined this way
Why it isn't defined the same way segments are?
how can the definition of an object include the object itself in it?
and how can I understand this definition ?

 A: 
Why [an Intervalle] isn't defined the same way Segments are?

The trick is that and interval $I$ can be "unlimited", like $[0,+∞)$, while a segment $[a,b]$ cannot. 
A: Of course one can define intervals the same way segments are defined. For example, in the book

Munkres, James: Topology

one finds the following definiton in §14 The Order Topology:
If $X$ is a simply ordered set [see the definition on p.24], there is a standard topology for $X$, defined using the order relation. It is called the order topology; in this section, we consider it and study some of its properties.
Suppose that $X$ is a set having a simple order relation $<$. Given elements $a$ and $b$ of $X$ such that $a < b$, there are four subsets of $X$ that are called the intervals determined by $a$ and $b$. They are the following:
$\phantom{xxxxxxxxxxx} (a,b) = \{x \mid a <x < b \}$
$\phantom{xxxxxxxxxxx} (a,b]= \{x \mid a < x \le b \}$
$\phantom{xxxxxxxxxxx} [a,b)= \{x \mid a \le x < b \}$
$\phantom{xxxxxxxxxxx} [a,b]= \{x \mid a \le x \le b \}$
The notation used here is familiar to you already in the case where $X$ is the real line, but these are intervals in an arbitrary ordered set. A set of the first type is called an open interval in $X$, a set of the last type is called a closed interval in $X$, and sets of the second and third types are called half-open intervals. The use of the term "open" in this connection suggests that open intervals in $X$ should turn out to be open sets when
we put a topology on $X$. And so they will.
Note that Munkres only defines bounded intervals; we could add the definitions
$\phantom{xxxxxxxxxxx} (a,\infty) = \{x \mid a <x  \}$
$\phantom{xxxxxxxxxxx} [a,\infty) = \{x \mid a \le x \}$
$\phantom{xxxxxxxxxxx} (-\infty,b) = \{x \mid  x < b \}$
$\phantom{xxxxxxxxxxx} (-\infty,b] = \{x \mid  x \le b \}$
to obtain unbounded intervals for all $a, b \in X$. Here $\infty$ and $-\infty$ are only formal expressions and no elements of $X$. We can moreover define $(-\infty,\infty) = X$.
Note that $(a,\infty)$ or $(-\infty,b)$ may be empty for certain $a,b$; this happens if $X$ has a smallest element $a$ or a biggest element $b$. Observe that $X = \mathbb R$ does not contain such elements.
There are quite a number of cases which have to be covered by the "segment-like" definition. So we see that the definition in your question is much shorter and more elegant.
Let us now consider $X = \mathbb R$ and compare your definition with Munkres's definition.
The definition of the segment $[a,b]$ works for any two $a,b$. That the segment $[a,b]$ is also defined for $b < a$ is just a formal (but useful)  issue. If $a < b$, your definition agrees with Munkres's definition of the closed interval. Munkres's definitions formally also work for $a \ge b$. This gives $[a,a] = \{a\}$, whereas all other intervals are empty. However, Munkres does not consider this case.
Note that the empty set and the singleton sets $\{a\}$ are intervals in the sense of your definition, but not intervals in the sense of Munkres. In the sequel let us only consider intervals which contain at least two distinct points.
We show that intervals in the sense of your question agree with intervals in the sense of Munkres.
It is fairly obvious that each interval in the sense of Munkres is an interval in the sense of your question, but we have to consider many cases to prove it formally. As an example consider $(a,b)$ with $a < b$. Take $x,y \in (a,b)$. W.lo.g. we may assume $x \le y$. Then if $z \in [x,y]$, we have $a < x \le z \le y < b$, thus $a < z < b$ which means $z \in (a,b)$.
The converse is a bit more complicated. Let $I$ be an interval in the sense of your definition which contains at least two distinct points.
Case 1. $I$ is bounded. This means that there exist $m, M \in \mathbb R$ such that $m \le x \le M$ for all $x \in I$.
Define $a = \inf I$ and $b =  \sup I$. Clearly $m \le a < b \le M$ and $a \le x \le b$ for all $x \in I$. In other words, $I \subset [a,b]$.

*

*If $a, b \in I$, then $I = [a,b]$: By definition, $[a,b] \subset I$.

*If $a \in I, b \notin I$, then $I = [a,b)$: Clearly $I \subset [a,b)$. Let $c \in [a,b)$. This means $a \le c < b$. By definition of $b$ there exists $y \in I$ such that $c < y$. Hence $c \in [a,y] \subset I$. Thus $[a,b) \subset I$.

*If $a \notin I, b \in I$, then $I = (a,b]$: Similar as 2.

*


*If $a, b \notin I$, then $I = (a,b)$: Also similar as 2. and 3.



Case 2. $I$ is unbounded.
Case 2a. $I$ has a lower bound $m$, but no upper bound.
Define $a = \inf I$. clearly $m \le a$ and $a \le x$ for all $x \in I$. In other words, $I \subset [a,\infty)$.

*

*If $a \in I$, then $I = [a,\infty)$: Let $c \in [a,\infty)$. Since $I$ does not have an upper bound, there is $y \in I$ such that $c < y$. We have
$c \in [a,y] \subset I$. Thus $[a,\infty) \subset I$.


*

*

*If $a \notin I$, then $I = (a,\infty)$: Clearly $I \subset (a,\infty)$.  Let $c \in (a,\infty)$. Since $I$ does not have an upper bound, there is $y \in I$ such that $c < y$. By definition of $a$ there exists $x \in I$ such that $a \le x < c$. We have $c \in [x,y] \subset I$. Thus $[a,\infty) \subset I$.



Case 2b. $I$ has a upper bound $M$, but no lower bound.
In this case we get as above $I = (-\infty,b]$ or $I = (-\infty,b)$.
Case 2c. $I$ has neither a lower nor an upper bound.
In this case we get $I = (-\infty,\infty) = \mathbb R$.
A: Following those definitions you will note that a segment is a special type of interval: it must be closed and bounded.  Closed and bounded subsets of $\mathbb R^n$ are important in analysis because they provide compactness (don't worry about what this means, if you never heard of compactness).
