Prove using strong induction that if G is connected and acyclic then G is also connected and has n-1 edges

Question

Prove that if G is connected and acyclic then G is connected and has n-1 edges.

My attempt at a solution, (I'm concerned there is an error in the inductive step):

Proceed by strong induction on n the number of edges.

Base Case: For $n=1$ we have an isolated vertex, which is connected and acyclic and we have $1-1 = 0 $ edges.

Inductive Step: Assume that the claim is true for all $1 \leq n \leq k$. That is every connected and acyclic graph with n vertices is connected and has n - 1 edges.

Inductive Hypothesis: We now prove the claim for $n = k+1$ vertices. Remove an arbitrary vertex $v \in V$ and its adjacent edges. There are now two possible cases, let $G^*$ be the graph obtained by removing vertex $v$. $G^*$ can either be connected or disconnected. In the case in which $G^*$ is connected, then we have k vertices and by the induction hypothesis we know the claim holds. In the case in which $G^*$ is disconnected, then we have two connected components. Let one component have q vertices and let the other component have w vertices such that $q + w = k$ and $1 \leq q,w < k$. By the inductive hypothesis we know that both of these connected components have $q -1$ and $w -1$ edges. If we add an edge to connect these two components we then have $q-1 + w-1 +1 = q + w -1 = k -1 $. edges. Therefore, by the principle of induction the statement is true.

Answer 1

It's simpler to remove an edge from $G$ rather than a vertex in the inductive case.

In the case in which $G^∗$ is connected, then we have $k$ vertices and by the induction hypothesis we know the claim holds.

You need to add back why $G$ has $k+1$ edges.

In the case in which $G^∗$ is disconnected, then we have two connected components.

This isn't true. Consider the following graph with the center vertex removed.