How can it be proved that each vertex can be at most in one strongly connected component in a directed graph? How can it be proved that each vertex in a directed graph will exactly be in at most one strongly connected component? I do not see it in the graph below which I think contains a couple of connected components and the one node in the middle seem to belong to two different connected components. Does the graph below contain one connected component or two connected components?

 A: Let a vertex $v$ be in two different strongly connected components $G_1$ and $G_2$. Then there is one vertex $v_1 \in G_1\setminus G_2$ and one vertex $v_2 \in G_2\setminus G_1$ so that $v$ is strongly connected to both of them. Therefore $v_1$ and $v_2$ are also strongly connected to eachother via $v$, and thus have to be part of the same strongly connected component. Thus a contradiction, and the assumption has to be false.

Edit after the question was updated:
So, in the image given in the question, the right diamond shaped side of the graph is strongly connected, and the left diamond is strongly connected, and the question is, are they part of the same strongly connected component, seing that the middle vertex is in both of them? (Correct me in the comments if this is not what you were asking)
The answer is yes, the whole graph is strongly connected. Pick any two vertices. If they are in the same half of the graph, just go around the diamond. If they are in different halves, go around the diamond until you get to the middle vertex, then go around the other diamond until you get to your other vertex.
A: We can show that belonging to the same strongly connected component is an equivalence relation.  The strongly connected components thus form the equivalence classes under this relation.  We can also show that in any equivalence relation no element belongs to two distinct equivalence classes.
