2. HINT:
let y be arbitrary in R. Consider open neighborhood $(y- \epsilon, y+\epsilon)$ and using the fact that there's at least 1 rational number between any two real numbers. This proof follows from topological version of dense set.
About the fact I claim above, It's good for future pure mathematician that you should prove before using.
For irrational case, it follows similarly and adjust something.
But if you hate topological version, you might use sequential version of dense set. For any arbitrary $a \in R$ , consider $ (a- \frac{1}{n}, a+\frac{1}{n}) $ for each n and applying the same fact.
5. HINT:
Let $ a \in R_d $ , consider open ball radius 1 ,say, B[a;1] = {a} to conclude that every subset is open. For proving closed, it's automatically from theorem that complement of open set is closed set.