Prove that any set in (X,d) is either closed or open. Let $X = \mathbb{R^2}$, $d(x,y) = |x| + |y|$ for $ x \neq y$. Prove that any set in (X,d) is either closed or open.

For the solution:

$ A \subset (\mathbb{R^2},d)$ for A to be open $ \forall a\in A\ \exists\epsilon\ B(a,\epsilon) \subset A$

By choosing $\epsilon$ as $|a|/2 $ $\ B(a,|a|/2) = \{a\}$ Thus A is open.
I can do the same logic for X\A but I do not know how to do either step or if A is such that the open... step.
Note that every $x \neq 0$ is an isolated point: $B(x, |x|)=\{x\}$. 

So if $0 \notin A$, $A$ is open. Otherwise the complement is open, so $A$ is closed.
First note that the sets $X$ and $\emptyset$ are both open and closed in a topological space $X$. Depending on your topology, every set in the topology might also be an open set, but in a metric space, this cannot happen. 

Consider a subset $A\subset X$. Let $A$ be open. We can show $A$ is not closed. 
Note since $A$ is open, this means $X\setminus A$ is closed. Since $A$ is an open set in a metric space, we can write it as an open ball $B_{\epsilon}(x)\subset X$ (since the topology is induced by the open balls). But then we have $X\setminus 
 B_{\epsilon}(x)=\{\forall x\in X: d(x,p)\geq \epsilon\}$ which is a closed set. We can apply a similar argument to show that $A$ cannot be open if $A\subset X$ is closed.