Consider the set $U = \{(x, y) \in \mathbb R^2: y > 0\}$. Working in the metric space $(\mathbb R^2, d_E)$, find open balls $B_1, B_2, B_3,\ldots$ with $U = \bigcup_{i \in \mathbb{N}} B_i$. Then explain why it is not possible to do this for the set $U = \{(x,y) \in \mathbb R^2 : y > 0\}$.
Should you approach using the open balls to find the boundary of the set?
Hint:
(I'm not sure about your notation, but I'll write $B_r(x,y)$ for the ball centred at $(x,y)$ with radius $r$.)
You don't need to find the boundary of the set: you just have to find a way to exactly cover the set with countably many open balls.
You're right in that the boundary is really where the problem is at though. For everything not "close" to the boundary, a bunch of balls like $S = \{B_1(x,y) : x,y \in \mathbb Z, y > 0\}$ will do: kind of like this, but with infinitely many.
$\hskip 2in$
For points "close" to the boundary, we can keep using balls of radius $1$, just for simplicity. All we really need is for them to get arbitrarily close together using only countably many balls. To do that, let $T_n = \left\{B_1\left(\frac{x}{n},1\right) : x \in \mathbb Z\right\}$ for $n \in \mathbb Z^+$. So $T_1$ would simply be the bottom row of balls in the above diagram. $T_2$ would look like:
$\hskip 2in$
And as $n$ grows large, these balls get closer and closer together. To make sure we have everything, we take the union of all these: $T = \bigcup_n T_n$. Note $T$ is a countable union of countable sets, which is itself a countable set of balls. Then $S \cup T$ should be the countable set of balls you're looking for. I'll leave it to you to prove that any $(x,y)$ with $0 < y < 1$ will be covered by some ball in $T$.
For the second part, a countable union of open sets is open. Can you show that $\tilde{U}$ is not open?
