This is a nice proof, and I believe it's basically correct. I had to guess what you meant at one point, though:
Then we proceed to add $e=(u,w)$ with $u \in K_i$ and $w \in K_j$ with $g(v)\geq K\geq j>i$ . This adds up to $g(v)-1$ edges to $G-v$.
This sounds like you're adding either one such $e$, or all of them, but from the structure of the proof it seems that what you mean is something more like:
Then we proceed to add, for each $i$ with $1\le i\lt K\le g(v)$, a new edge $e=(u,w)$ for some $u\in K_i$ and some $w\in K_{i+1}$, for a total of up to $g(v)-1$ new edges.
(Also you should perhaps introduce the notation $K_i$ for the components.)
I'll lay out an approach / hint in the first part, and then complete the proof in the second part of the post.
We can assume $k>1$.
If $G$ has a vertex of degree $k$, we are done. So we assume that the maximum degree of $G$ is some $\Delta \leq k-1$.
We let $u$ be a vertex of $G$, and assume for the sake of contradiction that $G$ does not contain any path of length $k$. In particular, this means that the diameter of $G$ is at most $k-1$ (since $G$ is connected and has no $k$-path), so every vertex of $G$ is within distance $k-1$ of $u$ (this is where connectivity is useful!).
If you just wanted a hint / way to approach the problem, stop reading here.
Now given the degree and distance constraints we have so far, let's see how many vertices $G$ can have.
We introduce the notation:
$$N_i(u) = \{v\in G : d(u,v) = i\}$$
The vertex $u$ has at most $(k-1)$ neighbours at distance $1$ from it (i.e., we have $|N_1(u)| \leq k-1$). Each of these in turn has at most $(k-2)$ neighbours besides $u$. So the set $N_2(u)$ of vertices distance $2$ from $u$ has $|N_2(u)| \leq (k-1)(k-2)$ (see the diagram).
In general, we get that $|N_i(u)| \leq (k-1)(k-2)^{i-1}$. So the total number of vertices $n$ is bounded above as follows:
$$n = |\{u\}| + |N_1(u)| + \dots + |N_{k-1}(u)|$$
And thus:
$$n \leq 1 + (k-1) + (k-1)(k-2) + \dots + (k-1)(k-2)^{k-2} = 1 + (k-1)\sum_{i=0}^{k-2}(k-2)^i$$
By using the formula for the sum of a geometric series, and a bit of inequality juggling, you can show that $n \leq k^k$. Thus the total possible number of edges that $G$ has is at most $\binom{n}{2} \leq \binom{k^k}{2}$. But this yields a contradiction, so $G$ must have a $k$-path.
Best Answer
Let $T$ be a spanning tree of $G$. Remove all the edges of $T$ from $G$. You will be left with a graph with $100$ vertices and $100$ edges which will necessarily contain a cycle (in general, a graph with $n$ vertices and $n$ edges contains a cycle). It's easy to see that removing this cycle from $G$ leaves it connected.