I am to find a set that is bounded but not closed nor compact. Here are my ideas. Please tell me if any of my logic is flawed. I thank you in advance.
Consider the set $A = (0,1)$ where $A \subset \mathbb{R}$ with the usual toplogy. This set is bounded since $\exists r > 0 \; \exists x \in \mathbb{R} : A \subset B_r(x).$Namely, the ball of radius 2 centered at $x = \frac{1}{2}$ contains $A,$ i.e., $(0,1) \subset B_2(\frac{1}{2}).$ $A' = (0,1)' = (-\infty,0] \cup [1, \infty)$ is not an open set, since $A' \notin \tau.$ Hence A not is closed. Since we know that $A$ is not closed, by the Heine-Borel theorem, we also know that $A$ cannot be compact.
You are right, but it could be simplified a bit more.
$A$ is an open ball of radius $\frac{1}{2}$ centered at $\frac{1}{2}$. So it is bounded. It is an open set, therefore it is not compact. An explicit example for an open cover that does not have a finite subcover would be the balls centered at $x_n$ with radius $r_n$, where $x_n$ is $\frac{1}{n}$ and $r_n$ is $\frac{1}{n+1}$.