(a) depends on what kind of world you live in. If there's only one person in the world the statement is false. So let's suppose there are at least two (and finitely many) people in the world. That is, our graph has vertices. For every vertex , there are possible values of : namely . Assume, for contradiction, that no two vertices have the same degree. Then, since there are vertices and possibilities for a degree, we can number the vertices such that the first vertex has degree 0, the second vertex has degree 1, the third vertex has degree 2,..., the -th vertex has degree . But now we see there is no graph with these degrees of vertices: since the first vertex has degree 0, the -th vertex can have degree at most , a contradiction.
(b) You have this correct. I wanted to add "Vertices of even degree don't play no role as to parity of SUM(deg(v))" but I guess it is too obvious.
(c) Suppose there are exactly two vertices of odd degree in and there's no path connecting them. Then the connected component of containing the vertex doesn't contain . So is a graph that contains exactly one vertex of odd degree (this is where we applied the theorem you mentioned). But this contradicts part (b).