What's wrong with this proof? Graph theory

Suppose we change the statement we're trying to prove: replace "connected graph $G=(V,E)$ with $|V|=|E|+1$" with "graph $G=(V,E)$ with no isolated vertices amd $|V|=|E|+1$". (Also, to give us a nice basis for the induction, let's just consider graphs with at least $2$ vertices.)

The "proof" seems to work the same as before. The only place we used connectedness was when we said "because we assume $G'$ is connected, this new edge must connect the new vertex to some vertex in $V$." But we can just as well say "because we assume $G'$ has no isolated vertices, this new edge must connect the new vertex to some vertex in $V$."

But the modified statement is false: the graph $G=K_3+K_2$ has $4$ edges and $5$ vertices, has no isolated vertices, and is not a tree. So what went wrong??

In a proof by induction on the number of vertices, going from $n=4$ to $n=5$ means that we assume that the statement holds for every $4$-vertex graph, and we have to prove it for every $5$-vertex graph. Note that we tacitly assumed that every $5$-vertex graph which satisfies the hypotheses can be obtained by adding a vertex to some $4$-vertex graph which satisfies the hypotheses. For the modified statement this is false; the graph $G=K_3+K_2$ can not be obtained by adding a vertex to a graph with $4$ vertices and $3$ edges and no isolates.

Going back to the original statement, a connected graph with $5$ vertices and $4$ edges can be obtained by adding a vertex and an edge to a connected graph with $4$ vertices and $3$ edges (because it is a tree, so it has a leaf, and removing a leaf gets us the desired $4$-vertex graph), but that needs to be shown.

To repair the proof, we could say: Let $G$ be any connected graph with $n+1$ vertices and $n$ edges. Since $G$ has more vertices than edges, its average degree is less than two, so it has a vertex $v$ of degree less than two. The degree can't be zero (since $G$ is connected), so $v$ has degree one. Then the graph $G'=G-v$ has $n$ vertices and $n-1$ edges, and $G'$ is connected, because you can't disconnect a connected graph by removing a vertex of degree one.