What do you think about my first simple direct proof? What mark/grade would you give me? Besides, I am curious about whether you like the style.
Theorem
Let $I = [a,b]$ be a non-empty closed interval. A difference between elements of $I$ is less than or equal to $b - a$.
Proof
Let $I = [a,b]$ be a non-empty closed interval. Let $x_1, x_2 \in I$. We consider the difference $d = x_1 - x_2$. It remains to prove that $d \le b - a$. We do so by deducing that there is a non-negative real number - which will be denoted by $\varepsilon\;$ - such that $d + \varepsilon = b - a$.
By the definitions of $I$ and $x_1$, we have $x_1 \le b$. Thus there is a non-negative real number $\varepsilon_1$ such that $x_1 = b - \varepsilon_1$.
Similarly, by the definitions of $I$ and $x_2$, we have $x_2 \ge a$. Thus there is a non-negative real number $\varepsilon_2$ such that $x_2 = a + \varepsilon_2$.
By the definition of $d$ and the results of the previous two paragraphs, the following holds: $d = b - \varepsilon_1 - (a + \varepsilon_2)$. Applying an appropriate equivalent transformation, we see that $$d = b - a - (\varepsilon_1 + \varepsilon_2)\text{.}$$ The parenthesized sum is a real number, since i) the addends are real numbers and ii) the set of real numbers is closed under addition. Moreover, the sum is a non-negative real number, since i) the addends are non-negative real numbers and ii) a sum of non-negative real numbers is non-negative. Introducing yet another symbol, we call this sum $\varepsilon$, and we have $d = b - a - \varepsilon$.
By another appropriate transformation, we deduce that $d + \varepsilon = b - a$. As $\varepsilon$ is a non-negative real number, we may state that $d \le b - a$. This is the statement that remained to prove. QED