G is a forest -> every connected subgraph of G is induced
Definition of a induced subgraph: for x,y in V(F), xy is in E(F) if and only if xy is in E(G)
Proof: Be F a connected subgraph of G. Suppose that F is not induced. Then there is a pair of vertices x and y such that there is a xy edge in E(G) but not in E(F). The contrary (xy in E(F) but no in E(G)) can't be true because F is a subgraph.
The edge xy is a unique path between vertices x and y in G (because G is a forest). So since F does not have this edge, it must be disconnected, contradicting our choose of F. Then it follows that F is induced.
I'm not convinced of the following: every connected subgraph of G is induced -> G is a forest.
It's probably simplest to consider the contrapositive: If $G$ is not a forest then there is a connected subgraph of $G$ that is not induced.
If $G$ is not a forest then it contains a cycle and you can look at connected subgraphs with the vertex set of the cycle.