This question is mainly to understand the meaning of my professor's correction to a proof of a theorem I gave during an oral examination.
The question was to show that $\text{diam}A = \text{diam}\bar{A}$ where $\text{diam}$ is the diameter of a subspace of a metric space and $\bar{A}$ is the closure of the subset $A$.
I proceded the following way:
Since $\bar{A} \supseteq A$, then we have $\text{diam}\bar{A}\geq \text{diam}A$. To prove the other inequality, we procede this way: $\forall x,y \in \bar{A}$ there exist $a,b \in A$ such that $d(x,a) < \epsilon$ and $d(y,b) < \epsilon$ with $\epsilon > 0$. Then we know that $$d(x,y) \leq d(x,a) + d(y,b) + d(a,b) < 2 \epsilon + d(a,b) \leq 2 \epsilon + \text{diam}A$$ and by the arbitrarity of $\epsilon$ the inequality follows.
My professor said that I coudn't do it this way and that, instead, I should do this:
$\forall x,y \in \bar{A}$ there exist $a_n,b_n \in A$ such that $d(x,a_n) < \frac{1}{n}$ and $d(y,b_n) < \frac{1}{n}$, then we would have $$d(x,y) \leq d(x,a_n) + d(y,b_n) + d(a_n,b_n) < \frac{2}{n} + d(a_n,b_n)$$, the next step would be to take the sup of it all this way: $$\sup_{x,y \in \bar{A}} d(x,y) \leq \frac{2}{n}+ \sup_{n \in \mathbb{N}} d(a_n,b_n)$$ from which the inequality follows because of the archimedean property.
My question is, how are these two proofs different? Why is mine wrong?