Prove: If a graph, $G =\left<V,E\right>$ is connected and $|E|=|V|$ then there's a cycle in $G$
I think this should be proved by induction. This is certainly holds for $n=1$ (a vertex with a loop).
Now, for the general case, I figure that every vertex in the graph should have degree $2$, which should help with completing the proof. But how exactly?
What I did so far:
The equality $$\sum\limits_{v\in V} d(v) = 2\left|E\right|$$ is true for any graph, and $|E|=|V|$ from the question assumption, so $$\sum\limits_{v\in V} d(v) = 2\left|V\right|$$
Let $|E|=|V|=n+1$, and suppose that the result is proved for graphs up to size $n$.
If $G$ has no cycle, it is a tree (by definition). In a tree, there are vertices of degree $1$ (the leaves), so that let $v$ be a leaf, and let $e$ be the unique edge adjacent to $v$. Then the graph $(V\setminus\{v\}, E\setminus\{e\})$ has $n$ vertices and as many edges as vertices, and it is connected so by induction hypothesis, it has a cycle $e_1\cdots e_k$. Do you see how to finish?