Proving that any connected graph has a vertex whose removal results in a connected graph I want to prove that: for any simple, connected graph there is at least one node whose removal results in a connected graph.
Here is my proof:

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*Suppose that a graph $G$ is simple connected graph with $\delta(G)$ is the minimum degree of the vertices. If $\delta(G)=1$ then there is a node with one edge, and removing this node with its edge keeps $G$ connected


*If $\delta(G) \ge 2$, then $G$ cannot be a tree, and we must have a cycle in the graph $G$, as every tree has at least two leaves which means $\delta(G) =1$ for trees. Now suppose that the cycle is defined as $C= (v_0,v_1,\cdots,v_k,v_0)$. Then, removing the node $v_i$ from $C$ where $deg(v_i)$ is the lowest degree of the nodes in this cycle will certainly keep the graph connected.
This concludes the proof.
Is this a correct and complete proof?
 A: No. You misunderstood the definition of 2-connected. This says that you can't remove a vertex such that the graph becomes unconnected. You have merely proven that there exists a vertex that can be removed such that the graph remains connected.
The claim is not true either. For this, consider the 3-path. 
A: Not sure I am clear on your proof.  it is not true that removing a minimal valence node from a cycle keeps the graph connected:  think of a "benzene molecule", a graph consisting of a cycle and one node hanging off each cycle node.  
To prove the statement you want:  choose a node P at random.  For any node V define L(V) to be the length of a minimal path from P to V.  Then you can remove any node, W, with maximal L(W).  (if Q is any other node then the shortest path from P to Q can't go through W, as that would make L(Q) > L(W))
A: Here is an alternative and easy proof:
Assume that $G$ is connected. Take the longest path $P=v_1v_2...v_l$ in $G$. Since $P$ is the longest path, all neighbors of $v_l$ is on the path $P$. Removing $v_l$ from $G$ does not disconnect the graph because any vertex in $G-\{v_l\}$ is either on $P'=v_1v_2\ldots v_{l-1}$ or connected to a vertex on $P'$.
