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 $2$).