How do I show $f$ is bounded. Let $B$ be a closed unit ball in $\mathbb{R}^2$ with respect to the usual metric; $\rho$ be the metric defined by
$$\rho(x,y)=\begin{cases} |x-y|, & \text{ if } x \text{ and } y \text{ pass through the origin,} \\
|x|+|y|,& \text{otherwise}.\end{cases}
$$
Suppose the $f: B \rightarrow \mathbb{R}$ is uniformly continuous on $B$ with respect to metric $\rho$ on $B$, and usual metric on $\mathbb{R}$.
Prove that $f$ is bounded.
I tried to prove by contradiction, and I am able to handle this question in usual metric space. Can anyone suggest me, how do I solve with respect to the metric $\rho$?
 A: Take $\delta$ s.t. for all $x, y$, if $\rho(x, y) < \delta$ then $|f(x) - f(y)| < 1$.
Then by induction on $n$, we can show that for all $n$, for all $x$ s.t. $\rho(x, 0) < n\delta$, $|f(x) - f(0)| \leq n$.
Base case: $n = 1$. This is immediate.
Inductive step: suppose $\rho(x, 0) < (n + 1) \delta$ and $|f(x) - f(0)| > n + 1$. Then clearly $|f(x) - f(0)| > n$; then $\rho(x, 0) \geq n \delta$. That is, $|x| \geq n \delta$. Then consider $x' = x(1 - \frac{\delta}{|x|})$. We have $\rho(x', 0) = |x'| = |x|(1 - \frac{\delta}{|x|}) = |x| - \delta < n \delta$. Then $|f(x') - f(0)| < n \delta$. And since $\rho(x, x') = \delta$, we have $|f(x) - f(x')| < \delta$. Then $|f(x) - f(0)| \leq |f(x) - f(x')| + |f(x') - f(0)| < n \delta + \delta = (n + 1) \delta$. Contradiction.
Then in particular, take $n$ s.t. $n\delta > 1$. Then for all $x$, we have $\rho(x, 0) = |x| \leq 1 < n \delta$. Then for all $x$, $|f(x) - f(0)| < n \delta$. Then $|f(x)| \leq |f(x) - f(0)| + |f(0)| \leq n \delta + |f(0)|$.
So $n\delta + |f(0)|$ is an upper bound on $f$.
