Let $f,g:\mathbb{R}^2\times\mathbb{R}^2\rightarrow \mathbb{R}$ with $$f(x,y)=\frac{|x_1-y_1|}{1+|x_1-y_1|}+\frac{|x_2-y_2|}{1+|x_2-y_2|}$$ and $$g(x,y)=\begin{cases}0 & x=y \\ 1& x\neq y\end{cases}$$
I have shown that these two functions are metric, so $(\mathbb{R}^2,f)$ and $(\mathbb{R}^2,g)$ are metric spaces.
Now I want to draw the balls $B_1(0,1):=\{x\in \mathbb{R}^2: f(x,0)<1\}$ and $B_2(0,1):=\{x\in \mathbb{R}^2: g(x,0)<1\}$.
We have that $f(x,0)<1\Rightarrow \frac{|x_1|}{1+|x_1|}+\frac{|x_2|}{1+|x_2|}<1$, right?
How do we continue? For $B_2$ we have that $g(x,0)<1\Rightarrow \begin{cases}0 & x=0 \\ 1& x\neq 1\end{cases}<1$. What do we get from that?
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EDIT:
The graph of $\frac{x}{1+x}+\frac{y}{1+y}<1$ with $x,y>0$ is the following:
In the first case, sketching $f(x,0) = 1$ might help. In order to do that, it might help to sketch ${x \over 1+x} + {y \over 1+y}$ for $x,y \geq 0$ first, and generalise. From here, a sketch of $B_1(0,1)$ should be much easier.
In the second case, it looks like you have an error in your last line - I'm not sure if this is a typo or a genuine mistake but double check it nonetheless. In this case, what does $x$ have to be to get $g(x,0) < 1$?