Considering the metric in $\mathbb{R}^2$: \begin{equation} d_{a}=\begin{cases} \lVert x-y \rVert & \text{if x and y lie on the same line through the origin}, \\\ \lVert x \rVert + \lVert y \rVert & \text{otherwise} \end{cases} \end{equation} Show that $A=${$x \in \mathbb{R}^2: d_a(x,0) \leq 1$} is not compact.
I want to showing that either it is not bounded or that it is not closed. However, it is clearly bounded because $A \subset ${$x \in \mathbb{R}^2: d_a(x,0) \leq 1+\epsilon$}. But to show it is not closed I am having trouble, because the only convergen sequences are those that go along the same line of the origin, and thus we are working with the usual metric and the limits are all in the closed unit ball. I also tried to prove that $A^C$ is not open, but without luck because balls outside $A$ with a radius smaller than $1$ are simply straight lines with direction to the origin and you can find an $\epsilon$ such that $B(x,\epsilon) \subset A^C$ so I am getting that $A$ is in fact compact but that is not right.
An alternative way to see it is by using the fact that in a metric space, compactness and sequential compactness are equivalent. To do so, consider the sequence $a_n = \left(\cos \frac{1}{n}, \sin \frac{1}{n}\right)$. First show that $a_n \in A$ and then notice that $d_a(a_n, a_m) = 2$ for all $m\neq n$ which implies that $a_n$ cannot have a convergent subsequence (since any convergent subsequence would be Cauchy).