On two different proofs 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?
 A: Just to elaborate on some of the comments: looking at your argument, you can see that
there is an implicit step which it might be good to make explicit:
your last displayed equation reads
$$d(x,y) \leq \cdots \leq 2\epsilon + diam(A).$$
Now the implict step is that you want to take a sup over all pairs $x,y$ of elements
of $\overline{A}$ to conclude that
$$diam(\overline{A}) \leq 2\epsilon + diam(A).$$
Writing out this step, we see that $x$ and $y$ have disappeared, but that the quantity
$\epsilon$ persists.  And now, for this to be valid, we see that it is important to note
that $\epsilon$ can be fixed independenty of $x$ and $y$.  Luckily, this is possible: if we
fix $\epsilon > 0$ in advance, then for any pair $x,y$ of points in $\overline{A}$, we can find $a$ and $b$ in $A$ within distance $\epsilon$ of them, as you note.  But it is important to state that $\epsilon$ is fixed in advance.
The main reason for being pedantic about this is that in other contexts, one may try to make similar arguments in which it turns out that $\epsilon$ can't be fixed in advance.  This kind of issue (whether a certain quantity (in your case, the quantity $\epsilon$) can be chosen uniformly  in terms of some other quantities (in your case, $x$ and $y$) 
comes up constantly in analysis, which is why people make a big deal about it in introductory courses.
Final remark: the last step of your professor's argument is misstated (as was also noted).
One should add one more step to the second-to-last line, to conclude that
$$d(x,y) \leq \ldots \leq \frac{2}{n} + d(a_n,b_n) \leq \frac{2}{n} + diam(A)$$
(this is the same as the implicit step in your argument which I mentioned above),
and then pass to sups, to find that
$$\sup_{x,y} d(x,y) \leq \frac{2}{n} + diam(A).$$
In the end, the only differences between this (corrected) argument and yours are:
(a) using $1/n$ instead of $\epsilon$, which is just a matter of taste and style;
(b) making the implicit step explicit; and
(c) making it explicit that $n$ is fixed in advance of considering any particular
pair $x$ and $y$ (which is pedantic, but as I already remarked, important).
