1. Correct way to answer calculus question

Heres a question from a Foundations of Calculus past paper and i was wondering whether the answer given is really needed...

Q: Prove That Sup A = 10, where $A = (-\infty ,10)$

A: Since A is bounded, $\alpha$ = sup A exists by the least-upper bound axiom.
Since $\gamma = 10$ is an upper bound for A, $\alpha \leq 10$ by property 2 (Properties at end of question).
We want to show that $\alpha$ cannot be less than 10, which will imply $\alpha = 10$
Assume that $\alpha < 10$.

Let $x = \frac{1}{2}(10 + \alpha)$.

Then $x < \frac{1}{2}(10 + 10) = 10$ so $x \in A$.
On the other hand $\alpha = \frac{1}{2}(\alpha + \alpha) < \frac{1}{2}(10 + \alpha) = x$.

Therefore, $\alpha$ is not an upper bound for A, which contradicts property 1. Absurd.

Therefore, $\alpha \geq 10$ so $\alpha = 10$.

Properties: A real number $\alpha$ is the supremum of a non-empty set of real numbers A if and only if...
1; $x \leq \alpha$ for all $x \in A$; and
2; if $x \leq \gamma$ for all $x \in A$, then $\alpha \leq \gamma$

And heres my thought... Cant you just use the Archimedian Property of Rational Numbers to say that...

Since A is bounded, $\alpha$ = sup A exists by the least-upper bound axiom.
Since $\gamma = 10$ is an upper bound for A, $\alpha \leq 10$ by property 2.
We want to show that $\alpha$ cannot be less than 10, which will imply $\alpha = 10$
Assume that $\alpha < 10$.

By the A.P.O.R.N there exists an x s.t. $\alpha < x < \gamma$. Therefore, $\alpha$ is not an upper bound. Absurd... etc...

Since A is bounded, $\alpha$ = sup A exists by the least-upper bound axiom.
Since $\gamma = 10$ is an upper bound for A, $\alpha \leq 10$ by property 2.
We want to show that $\alpha$ cannot be less than 10, which will imply $\alpha = 10$
Assume that $\alpha < 10$.
By the A.P.O.R.N there exists an x s.t. $\alpha < x < \gamma$. Therefore, $\alpha$ is not an upper bound. Absurd... etc...