Why is an open ball in $\mathbb R^n$ not compact? By definition of compactness, an open cover of an open ball in $\mathbb R^2$ always has a collection of subcovers that cover the ball. But why is a  open ball not compact?

Theorem A compact subset of a Hausdorff space is closed.
Proof. Let $C$ be a compact subset of the Hausdorff space $X$. Let $x\in X\setminus C$. For every $c\in C$, choose an open set $U_c$ such that $c\in U_c$ and an open set $V_c$ such that $x\in V_c$ and $U_c\cap V_c=\emptyset$, which is possible because $X$ is Hausdorff. The family $(V_c)_{c\in C}$ is an open cover of $C$, so it admits a finite subcover; hence we have
$$
C\subseteq U_{c_1}\cup U_{c_2}\cup\dots\cup U_{c_n}
$$
for some $c_1,c_2,\dots,c_n\in C$. Now, setting
$$
V=V_{c_1}\cap V_{c_2}\cap\dots\cap V_{c_n}
$$
we have that $V$ is an open neighborhood of $x$ and $V\cap C=\emptyset$. QED
Since an open ball in $\mathbb{R}^n$ is not closed (because $\mathbb{R}^n$ is connected), it can't be compact.
