Let $D$ be the set of all finite decimals. I wish to show that $$\sup\{a \in D: a^2 \leq 3\} = \sqrt{3}.$$ Is the following a valid proof of the statement? It suffices to show that, $$\forall \epsilon > 0, \exists a \in D \text{ s.t.} \sqrt{3} - \epsilon < a.$$ However, knowing $a \in D$, so that $a \leq \sqrt{3}$, the proposition may be written as $$\forall \epsilon > 0, \exists a \in D \text{ s.t.} \sqrt{3} - \epsilon < a \leq \sqrt{3}.$$ So, let $\epsilon > 0$. Owing to the density of finite rationals in the reals (proven, e.g., here), we are guaranteed $a \in D$ for which $$\sqrt{3} - \epsilon < a \leq \sqrt{3}.$$
Is the proof really so simple, or am I missing something?