Consider $\mathbb{R}^2$ with the metric: $$d((x_1,y_1),(x_2,y_2))= \begin{cases} |y_1-y_2| \text{ if } x_1=x_2 \\ 1+|y_1-y_2| \text{ if } x_1 \neq x_2 \end{cases}$$
Show that $E= \{(x,0) : -1 \leq x \leq 1\}$ is not compact.
I wanted to show this two different ways - the usual way of finding a finite subcover and then by showing its now totally bounded.
My problem here is trying to understand what an open cover would even look like. In the usual space we just draw disks (or balls in $\mathbb{R^n}$). In this metric space are all open sets just bars of infinite width and of height of at least $|y_2-y_1|$?