$\omega(G) := \text{Number of connected components of $G$}$
It is true that $G$ is a forest if and only if $|E(G)| = |V(G)| - \omega(G)$, but here I'll try to prove only one direction:
$|E(G)| = |V(G)| - \omega(G) \Rightarrow G$ is a forest
Proof: The contrapositive form is
$G$ is not a forest $\Rightarrow |E(G)| \neq |V(G)| - \omega(G)$
Suppose that $G$ is not a forest and that $|E(G)| = |V(G)| - \omega(G)$.
Can I take, for example, $K_3$ (connected and cyclic graph - not a forest) and show that
$|E(K_3)| = |V(K_3)| - \omega(K_3)$, i.e. $(3 = 3 - 1)$ is false,
so it must be true that $G$ is not a forest $\Rightarrow |E(G)| \neq |V(G)| - \omega(G)$?
I don't know why I'm struggling to see whether this wrong or not; I suspect I'm having a rather simple misconception regarding proofs. I know I'm supposed to show that it is valid for every graph that isn't a forest while I'm using an example (using a counterexample to show a contradiction) in my proof.
However, "$G$ is not a forest and that $|E(G)| = |V(G)| - \omega(G)$" can't be true because I've shown a G that isn't a forest such that $|E(G)| = |V(G)| - \omega(G)$ isn't true. Also, I have either an equality or an inequality ($\neq$), so there isn't another option here.
Still, there may be forests where the equality holds and others where the inequality holds. In this case, neither conditional is true. Possibly the fact that I know beforehand that the inequality holds is causing me to use a sloppy proof. Can someone enlighten me where I messed up or why what I did is valid?
In some sense, this is the raven paradox. That is, if you are trying to prove "All ravens are black", pointing out a black raven is evidence that supports that hypothesis. But is it also evidence to point out my blue shirt, which does satisfy the contrapositive statement that all non-black things are not ravens?
While this may be an interesting discussion for philosophers, it doesn't have any impact on the logic-driven fields of mathematics. Adding an example to a proof may have (very) limited utility in providing motivation for taking an argument in a certain direction or clarifying an algorithm, it has no value from the perspective of proving the statement is true for an infinitely-large domain.
A more productive approach would be to work in that each connected component of a forest is a tree. A connected component $H$ must satisfy $\epsilon(H)\geq\nu(H)-1$ or else it wouldn't be connected. From that, you might work from the contrapositive. If $G$ is not a forest, then it contains a cycle, so at least one of its components satisfies $\epsilon(H)>\nu(H)-1$, and therefore all of its components together must satisfy $\epsilon>\nu-\omega$.