Prove that any connected sub set of a DAG is a DAG. Intuitively I know this is true and I am interested in honing my proving prowess. If anyone could provide a know proof or their own I would absolutely love to see a formal answer this question.
(DAG = Directional acyclic Graph )
There seems to be 2 criteria here. First is the sub set acyclic?
Yes because it cannot contain a cycle that doesn't exist in the super-set. 
Is it directional?
Yes because a directional graph does not contain directionless edges so by transitivity if A is in B and C is not in B, C cannot be in A.
Taking the question one step forward can we prove that any connected subset of a tree is itself a tree? 
Again intuitively it is obvious but I don't have the practice in abstract to really formalize my thinking. 
 A: Any subgraph of a DAG, no matter whether it is connected or not, is itself a DAG.

There seems to be 2 criteria here. First is the sub set acyclic? Yes because it cannot contain a cycle that doesn't exist in the super-set. 

This is a good proof.

Is it directional? Yes because a directional graph does not contain directionless edges so by transitivity if A is in B and C is not in B, C cannot be in A.

This looks a bit confused. Being directed is not a property of a graph that it really makes sense to check, is is simply a different kind of thing from an undirected graph. The edges in the original graph all, by definition, have direction, so any subgraph of it necessarily has directed edges too. This is hardly worthy of proof at all.
Your talk about "transitivity" here sounds like you're misunderstanding something, but it is not clear to me what that misunderstanding might be.

Taking the question one step forward can we prove that any connected subset of a tree is itself a tree?

This depends on how exactly you define a "tree". If it's a connected acyclic (undirected) graph, then (a) your argument that the subgraph is acyclic from before can be reused word-for-word, and (b) being connected is something you explicitly assume, so there's nothing more to prove.
