every element is a subset I need to find an example of set $X$ with five elements that every one of them is subset of this set $X$.
does the next set - $X=\{\{1\},\{2\},\{3\},\{4\},\{5\}\} $ answer this question?
Edit: A good example of a set with 4 elements will be the power set of $\{\varnothing,\{\varnothing\}\}$ ?
 A: As mentioned twice already in the other answers, the number $5$ according to the Von Neumann construction of the natural numbers fits the criteria that you are looking for.  It can be seen from the construction that for any natural number $n$ and natural number $m$ with $m<n$ that $m\in n$ as well as $m\subseteq n$.  (In fact, with this construction of the natural numbers, this is how the less than relation is defined in the first place!)
A simpler construction which is easier to read which also works is this:
$$A=\{\emptyset,\{\emptyset\},\{\{\emptyset\}\},\{\{\{\emptyset\}\}\},\{\{\{\{\emptyset\}\}\}\}\}$$
That is, the set contains the empty set enclosed in no brackets, one pair of brackets, two pairs of brackets, on up to four pairs of brackets.  Notice that $\emptyset$ is a subset of the above trivially since $\emptyset$ is a subset of every set.  Further, the empty set enclosed in a $k$ sets of brackets (with $1\leq k\leq 4$) is a subset because $A$ has as an element the empty set enclosed in $k-1$ sets of brackets as well.
A: what about $\{ \{\},\{\{\}\},\{ \{\}, \{\{\}\} \},\{\{\}, \{\{\}\},\{ \{\}, \{\{\}\} \}\},\{ \{\}, \{\{\}\},\{ \{\}, \{\{\}\} \},\{\{\}, \{\{\}\},\{ \{\}, \{\{\}\} \}\}\}$
this thing looks sick. 
A: The answer above is rather awkward.
What it's saying is, let $$S_0 = \emptyset$$ $$S_1 = S_0 \cup \lbrace S_0 \rbrace$$
$$S_2 = S_1 \cup \lbrace S_1 \rbrace$$
... and so on ...
Each of the sets has all the elements of the set before it, plus one new one. Moreover, every element of each of the sets is a subset of the set.
The awkward answer with all the braces is simply expanding out $S_5$.
