Cardinality of countable subsets of the continuum Assume the following result:

If $A$ is an index set with $\#A\leq\#\mathbb R$ and $\{X_{\alpha}\}_{\alpha\in A}$ is a family of sets such that $\#X_{\alpha}\leq\#\mathbb R$ for each $\alpha\in A$, then $\#(\bigcup_{\alpha\in A}X_{\alpha})\leq\#\mathbb R$.

Suppose that $E$ is a set that has the cardinality of the continuum. Is there any way to prove that the set of countable subsets of $E$ has the cardinality of the continuum, using the result above?
 A: Let $A$ be an element of the set of all the countable subsets of $\mathbb{R}$, which we shall denote simply by $\mathcal{P}_{\le\omega}(\mathbb{R})$
Assertion: $\mathcal{P}_{\le\omega}(\mathbb{R})\preccurlyeq\,^\omega\mathbb{R}$
We will identify $A$ with a $\omega$-sequence of elements of $\mathbb{R}$, that is, an element of $^\omega\mathbb{R}$, in the following way:


*

*If $A$ is infinite, let $\{a_n|\;n\in\omega\}$ be an enumeration of $A$. Then it is clear that the assignation $A\longmapsto(a_n)_{n\in\omega}$ is injective.

*If $A$ is finite, say $A=\{a_0,\dots,a_n\}$, then we define the $\omega$-sequence $(b_k)_{k\in\omega}$ defined by:
$$b_k=\begin{cases}
  a_k\qquad\qquad\text{if }k\le n \\
  a_n+1\qquad\text{ if }k>n
\end{cases}$$
And in this case, it is also clear that the correspondence $A\longmapsto(b_k)_{k\in\omega}$ is injective.
In any case, $\mathcal{P}_{\le\omega}(\mathbb{R})\preccurlyeq\,^\omega\mathbb{R}$
Now, on the one hand we have that $\mathbb{R}\preccurlyeq\mathcal{P}_{\le\omega}(\mathbb{R})$, because the function $r\in\mathbb{R}\longmapsto\{r\}\in\mathcal{P}_{\le\omega}(\mathbb{R})$ is obviously injective.
On the other hand, $\mathcal{P}_{\le\omega}(\mathbb{R})\preccurlyeq\mathbb{R}$, since $\;\mathcal{P}_{\le\omega}(\mathbb{R})\preccurlyeq\,^\omega\mathbb{R}\;$ and $\;^\omega\mathbb{R}\preccurlyeq\mathbb{R}$: in fact, $|^\omega\mathbb{R}|=\big(2^{\aleph_0}\big)^{\aleph_0}=2^{\aleph_0\times\aleph_0}=2^{\aleph_0}=|\mathbb{R}|$
From the Cantor-Bernstein theorem, we obtain that $|\mathcal{P}_{\le\omega}(\mathbb{R})|=|\mathbb{R}|=2^{\aleph_0}$
