Why is belonging not transitive? From Halmos's Naive Set Theory, section 1:

Observe, along the same lines, that inclusion is transitive, whereas belonging is not. Everyday examples, involving, for instance, super-organizations whose members are organizations, will readily occur to the interested reader.

Belonging seems transitive. Can someone explain?
 A: Consider the empty set $\phi,$ which has no members. And $x=\{\phi\}$ has one member (namely, $\phi$ is the only member of $x$). And let $y=\{x\}.$ 
So $\phi \in x$ and $x\in y.$
But $\phi\not\in y,$ because the only member of $y$ is $x,....$ and $x$ is not $\phi$ because $x$ has a member while $\phi$ has none.
A: My dog belongs to me and I belong to the American Mathematical Society... and so....
A: Based on your response to other answers, your question seems to be "Why do we define the belonging relation ($\in$) in a way that cares about the level of nesting of sets?", i.e. why do we say that $a \notin \{\{a\}\}$?
We could define a relation that ignores nesting, a sort of recursive belonging that works the way you seem to want belonging to work (with $a {\tt\  recursively-belongs-to\ } \{\{a\}\}$), so why don't we use that as the definition of $\in$?
One reason is that we want to be able to use sets for abstraction of concepts to let us ignore details we don't care about.  For example, the field of set theory eventually defines the natural numbers as sets: they define $0 = \{\}$, $1 = \{0\}$, and $2 = \{0,1\}$.  You shouldn't have to care about that right now, but what if someone asks you whether $0$ belonged to the set $\{1, 2, 3\}$?  With non-recursive $\in$, you can immediately answer "no", but what if we had used the recursive belonging relation?  In that case the answer would be "yes", because $0 \in 1$ and $1 \in \{1, 2, 3\}$.
Using the current (non-transitive) belonging relation means when we use sets to define composite objects, we can use the resulting things as mathematical objects with their own properties instead of having to care about the details of how they were built out of sets.
A: A vertex of a triangle belongs to the triangle. A triangle belongs to the set of all triangles. But, a vertex is not itself a triangle.
A: The difference between $\subset$ and $\in$ is that the former applies to expressions at the same level of nesting and the latter applies to expressions at one level of nesting apart from each other. So when you chain two $\in$'s together you get something at two levels of nesting, which is not in general comparable to a single $\in$. On the other hand, since $\subset$ doesn't change the level of nesting it doesn't have this problem.
This is the idea behind the example given in other answers of
$$
\varnothing\in \{\varnothing\}\in \{\{\varnothing\}\},\qquad \varnothing \not\in \{\{\varnothing\}\}.
$$
A: $42 \in \mathrm{Even} \in \mathcal{P}(\mathbb{Z})$ but $42 \not\in \mathcal{P}(\mathbb{Z})$ because 42 is not a set of integers.
$\text{Peter} \in \text{Humans} \in \text{Species}$ but $\text{Peter} \not\in \text{Species}$ because Peter is not a species.
A: Let $y=\{\emptyset\}$.  And $x=\{y\}$.  Then $\emptyset\in y$ and $y\in x$, but $\emptyset\not\in x$.
A: Maybe it helps if you draw sets slightly different from the usual way:

Unlike usually, the set is represented not by the loop, but by a dot connected to a loop (if we are not interested in the content of some specific set, I'll omit the loop). In the image above, we have a set $A$ (the labelled dot in the left, connected to a loop), which has three elements (the dots labelled $1$, $2$, $3$ inside the loop).
Now a subset looks like this:

You see, all dots encircled by the loop of $B$ are also encircled by the loop of $A$, indicating that this is indeed a subset of $A$. But the dot of $B$ is not encircled by  the loop of $A$, which means that $B$ is not an element of $A$.
Now let's add a subset of $B$:

You see, anything inside the circle of $C$ is also inside the circle of $B$, thus $C$ is a subset of $B$. But this necessarily means that anything in $C$ is also in $A$, thus $C$ is also a subset of $A$. That is, the subset relation is transitive.
Now let's look at elements instead:

You see, the dot of $B$ is inside the circle of $A$, so $B$ is an element of $A$. Also, the dot of $C$ is inside the circle of $B$, thus $C$ is in $B$. But the dot of $C$ is not in the circle of $A$, thus $C$ is not in $A$. Since this is obviously possible (I just gave an example), the element relation is not transitive.
Note however that this doesn't mean that you cannot find sets where the relation is transitive, just that generally it isn't. For example, take the following set:

Here $B$ is both element and subset of $A$, that is, the elements of $B$ (in this case, just $C$) are also elements of $A$. Such sets are actually quite important, as they are how natural numbers get defined in set theory (indeed, if $C$ is the empty set, which represents the number $0$, then in the above image $B$ represents the number $1$, and $A$ represents the number $2$).
A: Belonging is not transitive because we don't want it to be.
Suppose I have sets $A = \{1, 2\}$ and $B = \{3, 4\}$. Now imagine that we write, "Let $C = \{A, B\}$."
When we say "Let $C = \{A, B\}$," what we're saying is that we want $C$ to be a set with exactly two elements: one of the elements is $A$, and the other element is $B$. If we wanted $C$ to have any other elements, we would have said so!
We want to be able to define a set that contains $A$ but doesn't contain $1$. For that reason, when we design the rules of set theory, we choose to design them so that belonging is not transitive. 
A: Belonging means to be an element of a set, so that $x\in A$ means that $x$ is an element of the set $A$. You can visualise $A$ as a collection of points, and $x$ is one of these points. What you are thinking of, which is correct, is that if $A\subseteq B$, and $x\in A$, then $x\in B$ too. Here, $A\subseteq B$ means $A$ is a subset of $B$, which you can visualise as $B$ being a collection of points which includes all the points of $A$ and possibly more.
However, this is different from saying that $A$ belongs to $B$ unlike perhaps the colloquial meaning of the word. If we were to write $A\in B$, or $A$ belongs to $B$, then we mean as above that $A$ is one point whereby $B$ is a collection of such points including $A$. But here, $A$ is not one point, but a sub-collection of points in $B$. This is an important difference. So although it is true that $x\in A$ and $A\subseteq B$ implies $x\in B$, it is not true that $x\in A$ and $A\in B$ implies $x\in B$, which is the requirement for transitivity.
Indeed, there is also a difference between $\{x\}$ and $x$ for a point $x$. The former refers to the set containing only the point $x$, while the latter refers to the point $x$ itself. This is why $A=\{x\}$, $B=\{\{x\}\}$ is not a counterexample to the claim, for example.

Note: To be pedantic, it's not really rigorous to say for example that $x$ is a point, not a set; for how do you define a "point"? But hopefully the above helps you intuitively understand the difference better.
