Why is a set being compact equivalent to it being closed? The definition of compact in my book is
A set K is compact if every open cover of K has a finite subcover.
Take the set of all the points inside and on the unit circle (closed). Then any open cover bigger than the unit circle could have a finite subcover.
Similarly, couldn't the set of all the points inside the unit circle (open) follow similar logic? Any open cover of the circle (anything larger than the unit circle) could have a finite subcover
 A: What about the open cover $\{U_{\alpha}\}$, where $U_{\alpha}$ is the open  circle centered at the origin of radius $\alpha$? There is no finite subcover of this that covers the open unit circle.
A: Consider the following open cover of the interior of the unit circle.
$$ \mathscr{C} = \left\{ \ B( \mathbf{0}, r) \ \colon \ 0 < r < 1 \ \right\}, $$
where $B( \mathbf{0}, r)$ denotes the interior of the circle of radius $r$ centered at the origin.  Note that this cover has no finite sub-cover. 
In a topological space, not every compact set is closed. For example, let $X = \{ a, b, c \}$ with the indiscrete topology. Then every subset of $X$ is compact, but no non-empty, proper subset of $X$ is closed.
In a Hausdorff topological space, every compact set is also closed. And, since every metric space is also a Hausdorff space (with respect to a metric topology), so this fact also holds in metric spaces. 
However, not every closed set is compact. For example, the set of integers is closed but not compact in the set $\mathbb{R}$ of real numbers with the usual topology. 
A: In metric topology we need to have closed and bounded for compactness.
Also note that the definition of  compactness indicates that every open covering must have a finite subcovering.
Thus if you choose an open covering for your set with a finite subcovering, that does not make your set compact. 
