Is metric space without origin connected? Consider $\mathbb{R}^2$ with a distance function defined by  $d(x, y) = \|x - y\|$ (Euclidean distance) if $x$ and $y$ lie on a line through the origin. Otherwise define $d(x, y) = \|x\| + \|y\|$. We have that $M = (\mathbb{R}^2, d)$ is a metric space.
Question. Is $M \setminus \{(0, 0)\}$ connected?
 A: We first describe all open balls $B_r(x)$ for $x \in M$, $r > 0$. If $x = 0$, then $B_r(x)$ is just a Euclidean ball of radius $r$, but $r < \|x\|$, then $B_r(x)$ is the ball of Euclidean radius $\|x\| - r$ along with the open itnerval in the line containing $x$ and $0$ of radius $r$. If $r \ge \|x\|$, then $B_r(x)$ is just the open interval at $x$ of radius $r$ in the line through $x$ and $0$.
Back to the original problem. Consider the set$$S = \{(x, 0) : x > 0\} \subset \mathbb{R}^2 \setminus \{(0, 0)\}.$$We claim that it is both open and closed. To see that it is open, for any $s \in S$, note that $B_{\|x\|/2}(x)$ is contained in $S$, by the description given above. On the other hand, if $y \in \mathbb{R}^2 \setminus \{(0, 0)\}$ is a limit point of $S$, then for any $\delta > 0$, we must have that $B_\delta(y) \cap S \neq \emptyset$. Because $y \neq 0$, we may consider $B_{\|y\|/2}(y)$, which must be a small radial interval which does not contain the origin. The only way this can intersect $S$ is if it was completely contained in $S$, i.e. $y \in S$. Hence, $S$ is both open and closed.
