Let be $G=(V,E)$ a graph and let $U$ be a set containing circles in $G$. Is $(U, E)$ a matroid? Let be $G=(V,E)$ a graph and let be $U$ a set containing circles in $G$. Is $(U, E)$ a matroid? 
I draw an easy graph (undirected) and checked this property.  It does not hold for my example. 

My example was like a graph with nodes $V=\{1,2,3,4\}$ and edges $E=\{a,b,c,d,e\}$ with the circles $\{a,b,d\}$ , $\{a,b,c,e\}$ and $\{d,e,c\} $ and this one would be a matroid if  the empty set would not be missing. So the statement is not true or am I missing something? Any Help is much appreciated.
A subset $C$ of circles (which I will call cycles here) of a graph $G$ cannot form the independent sets of a matroid. The reason for this is that the empty set is always an independent set (see the first axiom here) of a matroid while it is always (trivially) acyclic in $G$. 

What happens if we append the empty set to $C$? Well, if $C' := C \cup \{\emptyset\}$ are the independent sets of a matroid then they form a simplicial complex (every subset of set in $C'$ is also in $C'$). In particular, for every edge $e$ in some cycle in $C'$ we have $\{e\} \in C'$. But since $C'$ is a collection of cycles of $G$ it follows that $e$ is a loop in $G$. But this implies that every cycle in $C'$ is a loop. So any matroid whose independent sets are of the form $C'$ is the trivial matroid (whose only basis is the empty set) together with some loops, that is, the uniform matroid $\mathcal{U}_{0,n}$ where $n$ is the number of loops. 

On the other hand, the cycles of any graph $G$ do yield a matroid (the cycle matroid of $G$) whose circuits are the cycles of $G$ and whose independent sets (respectively, bases) are the forests (spanning forests) of $G$.
From definition of matroid, 


  Every subset of member of independent set is also an independent set.


So, if $U$ are set of independent sets, then for $A \in U$, one has $\{e\} \subset A$ is also independent set, but it is obvious not true because $\{e\}$ contains no circle.