Discrete subspace of $\Bbb{R}$ is countable 
Show that any discrete subspace of $\Bbb{R}$ with usual topology is countable.

Let $U$ be a discrete subspace of $\Bbb{R}$, for each $x\in U$ choose an interval with rational endpoints $(a_x,b_x)$ containing $x$ and if $y$ is distinct from $x$ then choose distinct $(a_y,b_y)$. Now consider the map $(a_x,b_x)\mapsto x$ , this map is surjective and since collection of intervals with rational endpoints is countable range must be countable i.e. $U$ is countable.Is it correct?

Your own proof is almost there. You haven't really used the discreteness (it seems now that you claim this can be done for any subset of $\mathbb{R}$, but these are not all countable!).
So for each $x \in U$ pick $(a_x, b_x)$ with rational endpoints such that $(a_x , b_x) \cap U = \{x\}$. This uses that $\{x\} \subset U$ is open in $U$, which is the definition of being discrete. 
Now, if $(a_x, b_x) = (a_y, b_y)$, intersect with $U$ to get $\{x\} = \{y\}$ or $x=y$. Now every point of $U$ is guaranteed to have a different rational interval...
