Yes, a countable subset constructed like this will also be dense. Also, this works for an arbitrary metric space $X$ without isolated points instead of $\mathbb R^2$.
Consider $x\in X$ and $\epsilon>0$. Then there are infinitely many $k\in\mathbb N$ with $d(x,a_k)<\frac\epsilon2$. This is true even if $x\in\{a_k\mid k\in\mathbb N\}$ (why? Note: $x$ is not an isolated point). And among these infinitely many $k$ there are some $>\frac2\epsilon$. Then $d(x,b_k)\le d(x,a_k)+d(a_k,b_k)<\frac\epsilon2+\frac1k<\frac\epsilon2+\frac\epsilon2=\epsilon$.
As the problematic situation with isolated points in the generalization may be somewhat tricky, here's a propositionette that can be used:
Proposition: Let $X$ be a $T_1$ topological space, $A\subseteq X$ a dense subset, $a\in A$ not an isolated point of $X$. Then $A\setminus\{a\}$ is also dense in $X$.
Proof: Let $U\subset X$ be open and nonempty. Since $a$ is not isolated, $U\ne \{a\}$. Since $\{a\}$ is closed, the nonempty set $U\setminus\{a\}$ is open and hence intersects $A$. It follows that $U$ intersects $A\setminus\{a\}$$_\blacksquare$