1. F is a subgraph of every ST of G implies F contains no cycles

and

2. F contains no cycles implies F is a subgraph of every ST of G

You can try a proof by contradiction so in proof 1 it will be:
Assume F is a subgraph of every ST of G and F contains a cycle.
But if F contains a cycle then F is not a spanning tree. But then F cannot be
subgraph of a spanning tree, so 1 must be true.

Thanks for the boost. The first one is simple enough, but I'm just wondering about the second one...

For (2):
F is a subgraph of G. We assume that F is acyclic.

If F is disconnected, then F is not a ST and cannot be a subgraph of one.

If F is connected, and of course acyclic, it is a tree and a subgraph of G also. Since the spanning trees of G contain every vertex of G, F can either be contained in the spanning tree or it is in itself a spanning tree.

If F is connected, and acyclic, it is a tree and also a subgraph of G. Since every spanning tree of G contains all the vertices of G, F must be contained in the spanning trees of G, or is in itself a spanning tree.

If F is disconnected, but acyclic, all the vertices of F are still contained in the spanning trees of G, and F is a subgraph of the spanning trees of G.

Wait, I might have found a counterexample.
Look at this graph G. Now the graph on the right is a spanning tree of G. The one on the bottom is acyclic, however it is not a subgraph of the one on the right. Are you sure you had to prove it was a subgraph of every spanning tree or a spanning tree?

Okay, would it help knowing the spanning trees are isomorphic to one another? Just guessing? Or if we only say there exists some F, for example, the edge 2_3 in your picture, such that it is a subgraph of every spanning tree? (I mean even if there is just 1 vertex in common, K_1 is still a subgraph and a tree.)

Okay but since F doesn't have to be a spanning tree in itself, isn't it possible to find, say, just 1 vertex that will be a subgraph of every spanning tree? A complete graph of order 1 or even 2, that will be a subgraph of every spanning tree, if you understand what I'm saying?

Of course, but the premise was "F is an acyclic graph which is a subgraph of G". The graph on the right satisfies this property, however it is not a subgraph of the graph below, but the graph below IS a spanning graph of G. Where did you get this problem by the way?