Prove that $G$ is Eulerian if and only if every block of $G$ is Eulerian.
Attempt:
If G is Eulerian, then G has a partition into cycles edge-disjoint. I don't know how to apply it, though.
If every block is Eulerian, then every vertex in $G$ which is the edge-disjoint union of its blocks has even degree.
Yes, as you said by Veblen's theorem the edges can be partitioned into edge-disjoint edges.
Assume a block induces a graph with a vertex $v$ of odd degree, it follows that there must be one cycle for which $v$ contains only one of the two edges adjacent to it. If we add all of the vertices in that cycle the graph will still be $2$-connected ( proving this is slightly tricky, we have to consider the case in which we remove $v$ but it's doable). This contradicts the maximality of the block.