Is the set $\big\{(x,y) : x^2 + y^2 \leq 1\big\}$ compact in $\left(\mathbb{R}^2,\sigma\right)$? Define $\sigma$ by
$$\sigma \big( \left( x_1,y_1\right) ,\left( x_2,y_2\right) \big) =
  \begin{cases} 
      \hfill \big|{y_1 - y_2}\big|  \hfill & \text{ if $x_1 = x_2$} \\
      \hfill \big|{x_1 - x_2}\big| + \big|y_1\big|+\big|y_2\big| \hfill & \text{ if $x_1 \neq x_2$} \\
  \end{cases}
$$
I've already proved that $\sigma$ is a metric.  I've played around trying to sketch $B_\sigma \big(\left(0,0\right),1\big)$ and have come to the conclusion that it is a square rotated 45 degrees and not containing the edges.  It's vertices are at $(0,1), (0,-1), (-1,0), (1,0)$ and it is fully contained within the unit circle.
e.g. $\big<\big>$ if the "lines" were dotted.
Unfortunately, this has yet to produce an epiphany.
 A: Let $D = \{ x^2 + y^2 \le 1\}$. Consider the set 
$$C = \{x^2 + y^2 = 1,\ \ y \ge 1/2\}.$$
For each $(x, y)\in C$, let $B_{(x, y)}$ be the ball centered at $(x, y- \delta)\in D$ with radius $r = 2\delta$, where $\delta < 1/4$. First of all, $(x, y)$ is in $B_{(x, y)}$ as 
$$\sigma ((x, y), (x, y-\delta)) = \delta <2\delta =r.$$
On the other hand, if $(\bar x, \bar y) \in C\setminus \{(x, y)\}$, then $\bar x \neq x$ and so 
$$\sigma( (x, y-\delta), (\bar x, \bar y)) \ge |\bar y| \ge 1/2 > 2\delta = r. $$
Thus we have 
$$B_{(x, y)} \cap C = \{(x, y)\}.$$
Now for all $(x, y) \in D\setminus C$, let $\tilde B_{(x, y)}$ be a ball centered at $(x, y)$ so that $\tilde B_{(x, y)}$ does not contain $C$. Then the open cover
$$\left\{B_{(x, y)} : (x, y) \in C \right\} \cup \left\{\tilde B_{(x, y)} : (x, y) \in D\setminus C \right\} $$
of $D$ does not have a finite sub-cover as they must contain all $B_{(x, y)}$. Thus $D$ is not compact. 
A: $$\text {Let } D=\{(x,y) : x^2+y^2\leq 1\}.$$ $$\text {Let } c=\cup \{B_{\sigma}((x,0),3/5) : |x|\leq 1\}.$$ $$\text { For } |x|<1  \text { let }  V(x)=B_{\sigma}((x,3/5),3/5) \cup B_{\sigma}((x,-3/5),3/5).$$  $$\text {Let } F=\{V(x) : |x|< 1\}\text { and let } G=F\cup \{c\}.$$ Then $G$ is an open cover of $D$  . Observe that for $|x|\leq 3/5$ we have (1)... $(x,3/5)\in D \backslash c$, and (2)... $V(y)=\{y\}\times J$ where $J$ is the union of the real intervals $(0,6/5)$ and $(-6/5,0)$ , so $(x,3/5)\not \in V(y)$ unless $y=x$... The members of $F$ are pairwise disjoint, and each one has non-empty intersection with $D$. Conclusion: Any sub-cover of $G$ must include $V(x)$ for every $x\in [-3/5,3,5]$, so $G$ has no finite (or even countable) sub-cover 
