A graph with an equal number of edges and vertices contains a cycle as a subgraph I'm trying to prove

A graph with an equal number of edges and vertices contains a cycle as
  a subgraph

My attempt

Induction on the number of vertices:
Clearly holds for $n=3$
Assume true for all graphs on $\le n-1$ vertices.
For n:
Case 1: There is a vertex of degree 1, say $v$. Then $G = v \cup H$.
  By hypothesis, H contains a cycle, so G necessarily contains a cycle.
Case 2: No vertex of degree 1. So every vertex has degree $\ge 2$.
  Take minimum spanning tree of G, which contains no cycles and has at
  most $n-1$ edges. But our hypothesis says that any graph on $\le n-1$
  edges contains a cycle. Contradiction.

I'm not sure about my case 2, any verification would be helpful.
 A: Your proof is incorrect. A spanning tree exists if and only if the graph is connected. $deg(u) \ge 2 \not\implies G$ is connected. You haven't proved that the graph is connected.
Example:
Consider this graph with 6 nodes and 6 edges:
$1-2
, 2-3
, 3-1
, 4-5
, 5-6
, 6-1$
Hint 1:

 Try to use the inductive hypothesis on each connected component of the graph.

Hint 2:

 Use the fact that a connected component with $r$ number of nodes must have $\ge r-1$ edges.

Hint 3:

 Use the fact that the sum of the edges of all connected components must be equal to $n$, the number of nodes in the graph.

Hint 4:

 If there are $r \ge 1$ connected components, then the total number of edges in these connected components is $ m \ge n - r$. But we know that $m = n$ which implies that at least one connected component with $l$ number of nodes has more than $l-1$ edges. Now use the inductive hypothesis.

A: There are a couple of problems with your argument. First, as Banach Tarski pointed out, $G$ doesn’t contain a spanning tree unless it’s connected. Secondly, the non-existence of vertices of degree $1$ in $G$ does not imply that every vertex of $G$ has degree at least $2$: $G$ may have isolated vertices, vertices of degree $0$. Finally, your induction hypothesis is that each graph on fewer than $n$ vertices with as many edges as vertices contains a cycle; it does not say that a graph with $n-1$ edges has a cycle unless that graph has $n-1$ vertices, which yours does not. Thus, your attempt to apply your induction hypothesis is invalid. An induction argument is nonetheless possible; I suggest one possibility below.
HINT: Suppose first that $G$ is not connected, and let the components of $G$ be $G_1,\ldots,G_m$. For $k=1,\ldots,m$ let $n_k$ be the number of vertices of $G_k$ and $e_k$ the number of edges.


*

*Explain why $e_k\ge n_k-1$ for $k=1,\ldots,m$.  

*Show that there must be a $k\in\{1,\ldots,m\}$ such that $e_k\ge n_k$, and explain why that $G_k$ must contain a cycle.


Now suppose that $G$ is connected. Then $G$ has a spanning tree $T$. Let $e=uv$ be the edge of $G$ that is not in $T$, and explain how to construct a cycle in $G$ that contains the edge $e$.
