Assume that diameter of set is defined as $\displaystyle \text{diam}(A) = \sup_{x,y \in A} d(x,y).$
Prove that if $\text{diam}(A) < \infty$ , show that there exists $x',y' \in A^{-}$ s.t. $d(x,y) = \text{diam}(A)$, where $A^{-}$ is closure of $A$
Since $\text{diam}(A)$ is $\sup,$ there exists a sequence $(x_k,y_k)$ s.t. $\displaystyle \lim_{k \rightarrow \infty} d(x_k,y_k) = \text{diam}(A).$ Now since $\text{diam}(A) < \infty$ and $d(.,.)$ is continuous we have that $\displaystyle \lim_{k \rightarrow \infty} d(x_{k}, y_{k}) = d(\lim_{k \rightarrow \infty} x_k , \lim_{k \rightarrow \infty}y_k) = d(x_0,y_0)$ ,so $x_0,y_0 \in A^{-}$ s.t $d(x_0,y_0) = \text{diam}(A).$
And we are done.
Is there any thing wrong with above argument?
EDIT : assume we are dealing with euclidean space.
As @GreginGre mentioned, first we need to show that sequences actually converge. since we have $\sup < \infty$ we have that $x_k,y_k$ are bounded. now use Bolzano Weierstrass theorem to get convergent subsequence. let us rename convergent subsequence to $x_k,y_k$ again and it converges to $x_0,y_0$, now only part that requires justification is $\lim_{n \rightarrow \infty} d(x_k,y_k) = diam(A)$. This is true because subsequence converges to same limit.
Best Answer
You didn't prove that $(x_k)$ an $(y_k)$ both converge, and in fact you have examples where they do not converge.
For example, take $\mathbb{R}$ with classical distance, $A=[-1,1]$ and $x_k=(-1)^k, y_k=-(-1)^k$. Then $d(x_k,y_k)=2=diam(A)$ and $(x_k), (y_k)$ both diverge.
Here is some a hint to solve your exercise.
Hint. Show the set $\{(x,y)\in A\times A\mid d(x,y)=diam(A)\}$ is closed and bounded.