For $P(n+1)$, you clain "all vertices are of degree $1$". That is not true in trees at all. Also, you need to prove $P(n+1)$, while you have only "proven" $P(n)$.
What you must do is assume that $P(n)$ is true, i.e. "all trees with $n$ vertices have $n-1$" edges. Then, take any tree $G$ with $n+1$ vertices and prove it has $n$ edges.
To do that, here's a Hint:
Find one vertex of $G$ that has a degree of $1$ (a "leaf") and remove it.
I'm sorry, but your proof is incomplete; you fell into the so-called "induction trap".
For the induction step, you assume that any connected graph with $k$ vertices has at least $k-1$ edges; and you want to prove that any connected graph $G$ with $k+1$ vertices has at least $k$ edges. In trying to prove this, you assume that $G$ was obtained by adding a new vertex (which you confusingly call $k'$) to a connected graph with $k$ vertices. But you have not justified this assumption.
To justify your argument, you would have to show that every connected graph with $k+1$ vertices can be obtained by adding a vertex (and some edges) to some connected graph with $k$ vertices. In other words, you have to show that, given any connected graph with $k+1$ vertices, you can find a vertex whose deletion results in a connected graph. This is true, but requires a proof. (Well, it's true for finite graphs, which is what we're talking about. In a connected infinite graph, it's possible that deleting any vertex disconnects it.)
For an alternative approach, you can delete an arbitrary vertex $v$ from a $k$-vertex graph $G$, without worrying about whether $G-v$ is connected or not; you then apply the induction hypothesis to each connected component of $G-v$, however many there may be. In this approach, you have to use the style of induction where you prove the statement for $k$ assuming that it holds for all numbers smaller than $k$.
Best Answer
Take two vertices that are connected by one edge. Call them $(u,v)$ and the edge $e$. Collapse the vertices into one vertex. You obtain a graph $G'$ on one less vertex and one less edge. Apply the induction hypothesis, and convince yourself that if you have a cycle in $G'$ then the uncollapsed version must also have a cycle.