I am struggling to write a good, thorough proof. The proof is supposed to be logically rigorous, correct and complete (e.g. no hidden assumption). Moreover, style is important - the proof should be publishable as solution to an exercise, it should not sound or look like a proof written by a beginner (which I am, unfortunately). Finally, the proof should be digestible. Please comment my proof (below) and please note that I am not a native speaker; I always hope for people who can identify sentences that sound or look strange. Do you think that the proof is pedantic, childish, or robotic? Is something missing or should something be more explicit?
Please note: The degree of a vertex $v$ is denoted by $d(v)$.
Theorem
Let $G = (V,E)$ be a simple graph. If $d(v) \ge \frac{\left| V \right| - 1}{2}$ for all $v \in V$, then $G$ is connected.
Proof
We prove the theorem by contradiction. More precisely, we use the both ends method: in the context of the above implication, we assume the hypothesis and the negation of the conclusion. After stating our assumptions, it is sufficient to find a contradiction.
We state our assumptions as follows. Let $G = (V,E)$ be a simple graph such that $d(v) \ge \frac{n - 1}{2}$ for all $v \in V$, where $n = \left| V \right|$. For the sake of contradiction, negating the conclusion, we assume that $G$ is disconnected.
Hereafter, we conclude the proof by finding a contradiction. The contradition is a necessity to satisfy a unsatisfiable inequality. The inequality will be obtained by chaining another two inequalities (4) and (5). Before we do this, we establish the preliminary statements (1), (2), and (3).
Since $G$ is disconnected, there are distinct vertices $v_1,v_2$ that are not connected. By definition, there is no path between $v_1$ and $v_2$. Thus, necessarily, $v_1$ and $v_2$ are not adjacent (1).
Let $V_1$ be the set of nodes that are adjacent to $v_1$; let $V_2$ be the corresponding set for $v_2$. By statement (1) and the definition of a simple graph, the sets $V_1,V_2$ are subsets of the set $V \setminus \{v_1, v_2\}$, whose cardinality is denoted by $n'$. Since $v_1,v_2$ are distinct vertices, we see that $n' = n - 2$ holds (2).
Moreover, we see that $V_1$ and $V_2$ are disjoint (3); for if they were not disjoint, then $v_1$ and $v_2$ would be connected, since these vertices would be adjacent to a common vertex.
Hereafter, we deduce our unsatisfiable inequality: By statement (3), we have $\left| V_1 \right| + \left| V_2 \right| = \left| V_1 \cup V_2 \right|$; and due to the above subset relationship, we have $\left| V_1 \cup V_2 \right| \le n'$. Combining the two previous statements, we deduce $$\left| V_1 \right| + \left| V_2 \right| \le n' \text{ .}\quad\quad(4)$$
By our hypothesis, $\left| V_1 \right| \ge \frac{n - 1}{2}$; the same is true of $V_2$. Thus we may deduce the following: $$n - 1 \le \left| V_1 \right| + \left| V_2 \right|\quad\quad(5)$$ We chain inequalities (4) and (5); accordingly, we have $n - 1 \le n '$. We substitute, using statement (2), and see that $n - 1 \le n - 2$.
The proof looks good to me. Personally, I think that paragraphs 4 and 6, which cover statements (1) and (3), could be even more detailed. Other than that, the proof appears to be complete and even conforming to a punctilious style.