My question is can the following exist as a simple graph? Or is it not possible?
The graph is to have 7 vertices and 7 edges and is connected. Also, if any one of its edges is removed, it will not disconnect the graph.
My thought is that it does not exist (purely based on 'try and error'), but I am having trouble explaining the reason in words.
Appreciate the help!
On
If the graph is connected, it has atleast one spanning tree which will contain $6$ of the $7$ edges of the graph. The $7^{\text{th}}$ edge is a chord, meaning that the graph has only one (fundamental) circuit. Removal of the branches of the tree that are not part of this fundamental circuit will disconnect the graph. Thus, the only way to uphold the last condition is for all branches to participate in the fundamental circuit, giving you the circuit of length $7$.
As $\sum\limits_{v\in V}\deg(v) = 2|E|$ by the handshaking lemma, we get that the average degree of each vertex must be $2$.
As the removal of any edge will not disconnect the graph we get that $\delta(G)> 1$ as otherwise a degree $1$ vertex could be disconnected from the graph by removing it's sole edge.
These two facts together imply that the graph must in fact be $2$-regular. The only $2$-regular connected graphs possible are the cycle-graphs.
We learn then that the only graph possible which matches your stated conditions (connected, 7 vertices, 7 edges, and edge-connectivity at least $2$) is going to be the graph $C_7$.
(image from wolfram mathworld)