$x\in \{\{\{x\}\}\}$ or not? I wonder if we can we say $x\in \{\{\{x\}\}\}$?  
In one viewpoint the only element of $\{\{\{x\}\}\}$ is $\{\{x\}\}$. In the other viewpoint $x$ is in $\{\{\{x\}\}\}$, for example all people in Madrid are in Spain.
 A: If $x\in\{\{\{x\}\}\}$, then $x=\{\{x\}\}$, and we have a cycle $x\in\{x\}\in x$. In Zermelo-Fraenkel set theory, this behavior is forbidden by the foundation axiom. So there does not exist any set $x$ such that $x\in\{\{\{x\}\}\}$, no matter how complicated $x$ may be!
(One can study less-popular set theories without the foundation axiom, but I don't know anything about those.)
In your example, you can use the subset relation to model the relationship between Madrid and Spain, instead of the set membership relation. The set of people in Madrid is a subset of the set of people in Spain. If we identify a place with the set of people in that place, then we can simply say that Madrid is a subset of Spain. In that case, we wouldn't say that Madrid is also an element of Spain, because Madrid isn't a person.
A: $$x\in\{\{\{x\}\}\}\iff x = \{\{x\}\}$$
The latter is obviously not true for many sets.
Your example of "all people in Madrid are in Spain" is not good here, because, if $\{x\}$ is madrid, then spain would be some set which includes both $\{x\}$ and some other elements
A: If Mariano Rajoy is in Madrid, and Mariano Rajoy is in Spain and Madrid is in Spain, then expressing those facts as memberships of sets would look like

Madrid = { Rajoy, ... }
Spain = { Madrid, Rajoy, ... }

so this doesn't resemble your example at all.
A: As many here explained, this certainly don't have to be true, and assuming the axiom of regularity this can be disproved quite easily.
I will remark that it is consistent that the axiom of regularity fails, and that this situation does happen. Of course it would mean that $x=\{\{x\}\}$, but it is a plausible scenario (which, for example, follows from certain "anti-foundation" axioms).
But perhaps one thing missing from the answers here is the mention of a transitive closure. Given a set $X$ we define the transitive closure of $X$ as the set of all elements which can be "reached" by taking elements of elements and so on. The transitive closure of $X=\{\{\varnothing\}\}$ would be to take $X\cup\{\varnothing\}\cup\varnothing=\{\varnothing,\{\varnothing\}\}$, for example.
So while $x$ need not be an element of $\{\{\{x\}\}\}$, it is an element of the transitive closure of $\{\{\{x\}\}\}$. 
A: While there are some variants of set theory that allow, e.g., that a set is an element of itself ("Quine atoms") or such deeper level weirdnesses as in your question, the usual axiomatization of set theory (Zermelo-Frenkel with or without the Axiom of Choice) contains the Axiom of Foundation (or Regularity)

Axiom. For every nonempty set $a$, there exists $b\in a$ with $b\cap a=\emptyset$.

In your example, $a=\{\{\{x\}\}\}$, the only possibility for $b$ is $b=\{\{x\}\}$, and then $a\cap b=\emptyset$ which merely gives us that the only element of $a$ is not in $a$, i.e., $\{x\}\notin \{\{\{x\}\}\}$. In principle, this would still allow $x\in a$ - after all, $x\ne\{x\}$.
However, we can use a better $a$. The only way for $x\in\{\{\{x\}\}\}$ is that $x=\{\{x\}\}$ holds. Assume that this is the case.
Now let $$a=\{\,x,\{x\}\,\} = \{\,\{\{x\}\},\{x\}\,\}.$$
By the Axiom of Foundation there exists $b\in a$ with $b\cap a=\emptyset$.
Then either $b=\{x\}$ or $b=x=\{\{x\}\}$. But in the first case $x\in b\cap a$ and in the second case $\{x\}\in b\cap a$, contradiction. Hence the assumption that $x=\{\{x\}\}$ (or equivalently $x\in\{\{\{x\}\}\}$) could hold for some $x$ must be wrong.
A: there are a lot of good answers, but I want to add:  
In the set $A=\{Chris,Culter,5,Adam,\{\{x\}\},525\}$, the elements are, by definition, all things between $\{$ and $\}$. so they are exactly:
Chris ,  Culter,  5,   Adam, {{x}} , 525
so in $B=$$\{${{x}}$\}$ the only element between $\{$ and $\}$ is...?   
A: The answer to your question is no. Note that the set $ \{ \{ \{ x \} \} \} $ has one element, which is $ \{ \{ x \} \} $, not $x$. To say $ x \in \{ \{ \{ x \} \} \} $ means that $x$ is an element of $ \{ \{ \{ x \} \} \} $, implying that $ x = \{ \{ x \} \} $, which is not true if $x$ is a number. 
A: No, if it were true it would break many things, for example the set theoretic construction of the natural numbers:
$$ 0 = \emptyset \\
 1 = \{\emptyset\}\\
 2 = \{\{\emptyset\}\}\\
\vdots$$
