Prove that if $G$ is a $k$-edge-connected graph, then $G+K_1$ is $(k+1)$ -edge-connected (my proof) I have a proof that is different from the ones proposed previously and I want to see if my proof works as well. I want to do my proof by contradiction.
First, let $H= G +K_1$ and suppose for the sake of contradiction that it is not $(k+1)$-edge-connected. Let $X$ be a vertex-cut of $H$ where $|X|=k$. Also, let $v$ be the vertex of $K_1$ in $H$.
Case 1: If $v\notin\ X$, then since $v$ is adjacent to all other vertices in G, there is an edge from each vertex in $H-X$ to $v$. Thus, $H-X$ is connected, which is a contradiction.
Case 2: If $v\in\ X$, then $H-X= G- (X-{v})$. However since $G$ is $k$-edge-connected and $|X-v|=k-1$, then $G- (X-{v})$ is still connected because we have not removed enough vertices in order for sufficient edges to be removed to disconnect the graph. This is a contradiction. 
Then, $H$ is $(k+1)$-edge-connected. 
I was going to do an edge-cut instead but I am not sure how to fix it to make it work that way. 
For the first case, if I instead look at this as $X$ is an edge-cut can I say this is the case where no edge in the set is adjacent to $v$ , but then $G-X$ is still connected because of the remaining edges, there is an edge connecting every vertex of $G$ to $v$? 
 A: A $k$-edge-connected graph does not have to be $k$-connected (e.g. the Butterfly Graph is 2-edge-connected but not 2-connected (it has a cut vertex)). So you really have to argue with an edge-cut and not a vertex-cut.
Here is how to fix your proof:
First note that since $G$ is $k$-connected the minimum degree of $G$ is at least $k$ and thus $G$ has at least $k+1$ vertices.
Let $H = G + K_1$ and suppose for the sake of contradiction that $H$ is not $(k+1)$-edge-connected. Let $W$ be the vertex set of the copy of $G$ in $H$ and let $v$ be the vertex of $K_1$ in $H$. Further, let $F \subseteq E(H)$ be an edge-cut of $H$ where $|F| = k$. If $F$ contains any of the edges incident to $v$, then $H[W] - F$ is connected since $G$ is $k$-edge-connected. Further, since there are $k+1$ edges to $v$ and we are only removing $k$ edges at least one such edge remains and thus $H-F$ is connected. If $F$ does not contain any of the edges incident to $v$, then $H-F$ is clearly connected, since from every vertex there is a path (of length at most 2) going through $v$ to every other vertex. In either case $H-F$ is connected, thus we have a contradiction.
