For (a), I don't see how you argue that having each $B_r(x_0) \not\subset A$ implies that one of these is contained in the complement of $A$. It might be better to forget about contradiction: if $x = (x_1, x_2)$ is a point in the first quadrant, then let $r = \min(x_1, x_2)$ and show that $B_r(x) \subset A$. [Draw a picture!]
For (b), you'll have to say what your definitions are. When you say that a space is "discrete", that should mean that every subset is open. Perhaps you are starting from the standard discrete metric
$$
d(x, y) = \begin{cases}1, & x \neq y \\ 0, &x = y\end{cases};
$$
in that case, given $A \subset X$ you should be able to show that for each $x \in A$ we have $B_1(x) = \{x\}$, and this is certainly a subset of $A$.
Added later. Let me replace your questions with more questions.
(1) Show that for general $x = (x_1, x_2)$ and $y = (y_1, y_2)$ in $\mathbf R^2$ we have
$$
|y_1 - x_1|,\, |y_2 - x_2|\leq d(x, y).
$$
(2) Thus, in the setting of (a), for $y \in B_r(x)$ we have for example
$$
|y_1 - x_1| \leq d(x, y) < r \leq x_1
$$
and hence
$$
-x_1 < y_1 - x_1 < x_1 \quad \Rightarrow \quad 0 < y_1.
$$
For an approach even more basic than Andrew Salmon’s, let $\langle X,d\rangle$ be a metric space, and let $F$ be any finite subset of $X$. The empty set is closed by definition, so we might as well assume that $F\ne\varnothing$. Now suppose that $x\in X\setminus F$, and let $r_x=\min\{d(x,y):y\in F\}$. Then $r_x>0$ (why?); what can you say about $B(x,r_x)$, the open ball of radius $r_x$ centred at $x$?
Yes, a finite set in a metric space can be open. First, the empty set is always open. Other than that, though, it depends on the space. No finite, non-empty subset of $\Bbb R^n$ is open, for instance, for any $n\in\Bbb Z^+$. However, if $X$ is any set at all, the function $d:X\times X\to\Bbb R$ defined by
$$d(x,y)=\begin{cases}1,&\text{if }x\ne y\\0,&\text{if }x=y\end{cases}$$
is a metric, often called the discrete metric, and every subset of $X$ is open.
Best Answer
Singletons $\{a\}$ are closed sets, so their complements are open. For any set $A \subseteq X$ consider the following: $$A=\bigcap_{a \in A^{c}}\{a\}^c.$$ Observe that each $\{a\}^c$ is open and once you verify the set equality (which is not that difficult) you have $A$ as the intersection of open sets.