My goal is to prove that a closed and bounded set is compact. I have constructed a topological argument using the notion of diameter. My question is whether or not my proof is logically sound. Thank you.
Proof:
Let $A \subset \mathbb{R}^n$ with the usual metric be a closed and bounded set. I will connect the ideas of closed and boundedness using the diameter. Let $D$ be the diameter of $A$, defined as $ D = \max\{d(x,y) : x,y \in A\}$. Since $D$ is bounded, we know that the diameter of $A$ exists as a finite number, (otherwise there would not exist an open ball of radius $r$ that contains $A$, per the def of bounded) Since $D$ is closed, we know that $A$ contains the arguments of the maximum, $x^\star, y^\star$ (the elements that connect the diameter).
Suppose that $A$ is not compact. Then there exists an open cover $C$, such that any subcover does not cover $A$. Suppose that this open cover is defined as $C = \{B_r^n (x^\star)\}_{r \in \mathbb{R}}$, the collection of open balls of radius $r$, centered at $x^\star$. Clearly, as $C$ covers $\mathbb{R}^n$, it must cover $A$. However, any sub cover $C^\prime = \{B_{r_1}, B_{r_2}, \dotsc, B_{r_k}\}$ does not cover $A$ by assumption. Hence, for any $C^\prime$, there exists an $y \in A$, such that $y \notin \cup_{i = 1}^k B_{r_i}$. Suppose that $r^\star = \max \{r_1, r_2, \dotsc, r_k\}$. This implies that the element $y \in A$ is further away from $x^\star$ than the largest radius $r^\star$ is long (suppose $y$ is the furthest element), hence $y = y^\star$. We clearly see that $r^\star \leq D$. Since $r^\star$ can becomes arbitrarily large, this means that $D$ becomes arbitrarily large, which produces a contradiction, namely that $D$ is not a finite number. Therefore, $A$ must be compact.
QED
Best Answer
I don't think this proof checks out. You're first assuming that there is an open cover $C$ that has no (finite) subcover. Then, in the next step, you are assuming a very particular form for the cover $C$. Do you see why that is logically not allowed? There is no reason for the cover $C$ to consist of balls with the same center.
You might be able to remedy your proof by choosing $C$ in this way, but then you have to prove explicitly that it doesn't admit a finite subcover, something which I don't think is true.
I think you are better off by trying to reduce it to Heine-Borel in $\mathbb{R}$ by using the projection onto each of the components. Since $A$ is bounded, it fits inside of a closed rectangle $R$. Then, $R = [a_1, b_1] \times .... \times [a_n,b_n]$ is compact as the product of compact subsets of $\mathbb{R}$. Finally, $C$ is a closed subset of a compact, hence, it is compact itself.