Countable subsets of $\mathbb I$, bijections and $\mathbb R$

I have begun some studying of set theory at a level which could be described as "undergraduate level".

So there is some definition of cardinality and bijections between $$\mathbb N$$ and $$\mathbb Q$$, and uncountability of $$\mathbb R$$ and countability of $$\mathbb N$$ and $$\mathbb Q$$.

So, it is really a beginning of some possible further study, with just some basic concepts that I have grasped so far.

But, it seems legal to ask at this level, is there a bijection between the set of all countable subsets of $$\mathbb I$$ and $$\mathbb R$$?

Here $$\mathbb I$$ denotes the set of all irrational numbers, and if $$\mathbb I_C$$ denotes the set of all countable subsets of $$\mathbb I$$, this question is: Is there a bijection $$b: \mathbb I_C \to \mathbb R$$?

I would also like to see a constructed example of such bijection.

First, let's show that there is an injection $$f:\mathbb R\to\mathbb I_c$$ defined by $$f(x)=\begin{cases}\{x,x+1\}&x\notin\mathbb Q\\\{x+\pi\}&x\in\mathbb Q\end{cases}$$
Clearly, rationals and irrationals map to different subsets, since sets with two elements never equal sets with 1 element. Moreover, $$x+\pi$$ is irrational is $$x$$ is rational, and $$x+1$$ is irrational if $$x$$ is irrational. It's easy to see why this is injective.
Next, we note that there is an bijection $$h:\mathbb R^\mathbb N\to\mathbb R$$, since there is a standard bijection from $$2^\mathbb N\to\mathbb R$$ (binary representation), and hence we simply need a bijection from $$(2^\mathbb N)^\mathbb N=2^{\mathbb N\times\mathbb N}\to2^\mathbb N$$, but that's easy since there is a bijection from $$\mathbb N^2\to\mathbb N$$.
It's obvious that there is an injection from $$\mathbb I_c\to\mathbb R^\mathbb N$$, so we are done!