Let $G$ be a connected simple graph on $n$ vertices. Show that $G$ has exactly $n-1$ edges implies every edge of $G$ is a bridge.
The above statement is obviously true. Some textbooks even use this statements as the definition of a tree. However, my teacher suggests us to prove it directly as an exercise. It would be great if you can verify my proof below and/or give another direct proof of the statement.
Latest version:
We proceed by induction on $n$, the number of vertices of $G$. For the case $n=2$, $G$ has 2 vertices and 1 edge. It is clear that this edge is a bridge.
Assume $G$ has $n$ vertices and exactly $n-1$ edges. We claim that there must exist a vertex $v$ with $\text{deg}(v)=1$. If the claim is not correct, then $$n-1 = \text{number of edges} = \frac{1}{2} \times \text{sum of degree} \geq \frac{1}{2} \times 2n=n$$ which is a contradiction. Hence, such $v$ exists, with only one edge $\alpha$ incident to it. Now, on the removal of $\alpha$, $G$ is partitioned into two connected component $G_1$ and $G_2$. In particular, $G_1=\{v\}$ is a singleton. Hence, $\alpha$ is a bridge. By induction hypothesis, $G_2$ has $n-1$ vertices and exactly $n-2$ edges, thus every edge in $G_2$ is a bridge. Hence, every edge in $G$ is a bridge.
Wrong version:
We proceed by induction on $n$, the number of vertices of $G$. For the case $n=2$, $G$ has 2 vertices and 1 edge. It is clear that this edge is a bridge.
Assume that $G$ has $n-1$ vertices and exactly $n-2$ edges. By the induction hypothesis, these $n-2$ edges are bridges. Now, we choose an arbitrary vertex in G, join it to a newly-added vertex $v$ by an edge $\alpha$. It remains to show $\alpha$ is a bridge. Note that, on the removal of $\alpha$, the graph will become disconnected as there will be no edge incident to $v$. Hence, $\alpha$ is indeed a bridge and we are done.