The definition implies no such thing. There is no such thing as "implying" in definitions. The definitions state what they explicitly state, and nothing more and nothing less.
For example, if I say I am taller than $150$ centimeters, I do not imply that I am shorter than $151$ centimeters. The definition simply says that
No matter how you remove fewer than $k$ vertices, the graph remains connected.
This means that it should be easy for you to show that if a graph is $k$-connected, then it is also $k-1$ connected.
The reverse is of course not true, but your proof is not good. What you have "proven" is that a graph cannot be both $k$-connected and $k+1$ connected, which is not true (a counterexample of that is a full graph on $k+2$ vertices, for example)
There is also the concept of exact $k$-connectedness, and a graph is $k$-connected if it is $k$ connected, but it is not $k+1$ connected (and, consequently, the graph is $1,2,3,\dots,k$ connected, and is not $k+1,k+2,k+3\dots$-connected)
This is the concept used in your proof. You already know that the graph is $2$ connected (so it is also $1$ connected), now it can either be exactly $2$ connected or not. The proof shows that the graph cannot be exactly $2$ connected, so it must be $3$ connected.
The question is a little vague- for cut-edges, there is a precise answer, namely that the number of components increases by precisely one. This is fairly easy to see- if $e$ is a cut-edge, there will be one component of $G-e$ that contains one endpoint of $e$ and another component that contains the other end point. These are different components since $e$ is a cut-edge, and adding $e$ makes them into one component.
However, as you've already shown, you can't nail this down to an exact number for all cut-vertices, nor can it be determined just by the degree of the cut-vertex. I don't know of any generalized formula for how many components the deletion of a cut-vertex would add. I imagine the point of this question was for you to show that unlike cut-edges, cut-vertices do not add a predetermined number of components.
Best Answer
Ted's comment is correct, of course, but this is not always the easiest way to prove such an equivalence. I would just try to figure out which implications you can actually prove and hope that this is enough. (3)$\Rightarrow$(1) is indeed easy, and so is (1)$\Rightarrow$(3).
(2)$\Rightarrow$(3) is pretty clear as well. So now you know that (1) and (3) are equivalent and that (2) implies both of these properties.
It remains to show that (1) and (3) together imply (2). (Actually, you will only have to use (1).) I would suggest to go by induction on the number of vertices. If there are no cycles and the graph has at least 1 vertex, then there is a vertex of degree 1 (why?). Remove that and observe that the smaller graph that you obtain is connected and has no cycles.
Edit: I noticed that I contradicted myself somewhat: Clearly, I am actually providing hints to prove (1)$\Rightarrow$(2)$\Rightarrow$(3)$\Rightarrow$(1). But I still believe that a good strategy to prove the equivalence of more than two statements is to see which implications look easiest and then hope that you get enough for the equivalence.
Edit 2: For the implication (2)$\Rightarrow$(3) you should also use induction. If the graph has no more than $n-1$ edges, there must be a vertex of degree $\leq 1$. Since the graph is connected, that vertex has degree exactly $1$. Remove that vertex and adjacent edge and use the inductive hypothesis on the smaller graph. The base case is a two vertex graph joined by an edge.