I want to first note that a graph with $(|V|,|E|)=(4,6)$ does not disprove the inequality, since each side evaluates to $6$. Your example with $(|V|,|E|)=(5,7)$ cannot exist, due to the following proof.
Note that $$
|E(G)| \geqslant [\frac12(3|V(G)|+1)]\iff
\begin{array}{}
|E(G)| \geqslant \frac12(3|V(G)|+1)& \text{for odd }|V|\\
|E(G)| \geqslant \frac12(3|V(G)|) & \text{for even }|V|
\end{array}$$
Since $2|E(G)|=\sum_{v\in V}deg(v)$,
$$
\begin{array}{}
\sum_{v\in V}deg(v) \geqslant 3|V(G)|+1& \text{for odd }|V|\\
\sum_{v\in V}deg(v) \geqslant 3|V(G)| & \text{for even }|V|
\end{array}
$$
We will show that every vertex in a Hamiltonian connected graph must have at least degree $3$. Suppose (for the purpose of contradiction) that $G=(V,E)$ is Hamiltonian connected with $|V|\ge 4$ and that $v\in V$ has degree $2$. Say that $v$ has exactly neighbors $x,y$. Consider a Hamiltonian path from $x$ to $y$. Since $v$ has only two neighbors, then the first vertex in this Hamiltonian path must be $v$, or else $v$ will be unreachable before reaching $y$. However, if the vertex $v$ is second on the Hamiltonian path, then the only possible vertex to be third is $y$. But there must be at least $4$ vertices so this is not a Hamiltonian path. This is a contradiction, so the vertices of every Hamiltonian connected graph with $|V|\ge 4$ must have degree at least $3$.
For even $|V|$, since each vertex has degree at least $3$, $$
\sum_{v\in V}deg(v) \geqslant 3|V(G)|
$$
For odd $|V|$, since each vertex has degree at least $3$ and since the sum of all of the degrees cannot be odd, then $$
\sum_{v\in V}deg(v) \geqslant 3|V(G)|+1
$$
The proof overall has the right ideas but the presentation can be improved.
I have to show that there exists a path in the obtained graph that touches every vertex
How are you defining a path? If you require all the vertices in a path to be distinct, then this is not what you have to prove. It is even false. There is no path that touches every vertex in the tree with the edge-set 12, 23, 24.
What you want to prove is that there is a path between any two vertices in the contraction.
You continue with
Derived from a Theorem
We the readers of your proof do not know which Theorem you are referring to. A month, or a year from now if you re-read your proof, you might not know either!
Continuing with the same paragraph,
Derived from a Theorem, we know that every connected graph contains a spanning tree. Hence, we only need to prove that this lemma holds for spanning trees.
You are not making use of any special property of spanning trees in your proof. That entire paragraph is unnecessary because your proof is equally valid on all connected graphs.
Let e be an edge incident with vertices u and v
you could say "Let $e = uv$".
When you say
so that uev1...v is a path
it's not clear what this path is. You are using $u$, $e$ and $v$ in your path, so it's a mix of symbols that represent different things. You could say "Let $p$ be a path between $u$ and $v$. If $e\notin p$ then (...) and if $e\in p$ then (...).
When you say
then e is not on the path between vertices u and v in G'
we do not know what $G'$ is. You might have defined somewhere (e.g. in class) what $G'$ stands for, but we the readers do not know that. The standard notation is $G/e$.
The entire statement
If e is not on the path between the vertices u and v, then e is not on the path between vertices u and v in G'.
has nothing to do with $e$ not being on a certain path between $u$ and $v$. By the definition of contraction $e$ doesn't exist in $G/e$ so it can never be in any path in $G/e$. What you wanted to say is that if there is a path $p$ between $u$ and $v$ in $G$ such that $e$ is not on it, then $p$ remains a path between $u$ and $v$ in $G/e$.
Since a new vertex was created, there exists a new path including the new vertex w.
Why is this new path relevant?
Hence, G is connected.
That is our initial assumption, we know that. You meant $G/e$ is connected. How does that follow from the existence of this new path introduced above?
(In fact, this process can be repeated until only one vertex and no edges exist since edge never ends in loss of connectivity).
This observation is correct but is not relevant for the proof, so it shouldn't be included in the proof, but after it.
Best Answer
I tried to give as much details as possible :
Suppose that $G$ is hamilton-connected, i.e. for any pair of vertices $u$, $v$, there exists an Hamiltonian path from $u$ to $v$.
Suppose that in $G$, there exists a vertex $v$ of degree only 2, and call its neighbours $x$ and $y$.
If $G$ is hamiltonian, then it must exist an hamiltonian path from $x$ to $y$. In order to be hamiltonian, this path must include $v$ (as well as all other vertices of G, once only).
In the path $(x\ldots y$), suppose that $v$ is not in second position, i.e. the path is $(x\ldots av\ldots y)$. But then this means that $a$ is connected to $v$. $a$ can be only equal to $x$ or $y$ (the only neighbours of $v$). $a$ cannot by $y$ as an Hamiltonian path includes each vertex only once (and this one also ends by $y$). But $a$ cannot be $x$ too. This is a contradiction. Hence $v$ is in second position in the Hamiltonian path from $x$ to $y$. Therefore the path must be $(xv\ldots y)$.
But from $v$, the only vertex we can reach is $y$, therefore the path is exactly $(xvy)$
However we have $|V(G)|\geq 4$, therefore this path is not hamiltonian, (it would need to go throught every vertex in order to be hamiltonian). this is a contradiction.
Therefore there exist no hamiltonian path from $x$ to $y$, and then $G$ is not Hamilton-connected