prove that $|{\{X\subseteq\mathbb{N}|\ |\mathrm{X|={\aleph_{0}}}\}}| = 2^{\aleph_{0}}$

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I need to prove that:

$$|{\{X\subseteq\mathbb{N}|\ |\mathrm{X|={\aleph_{0}}}\}}| = 2^{\aleph_{0}}$$. it's allowed to use the fact that $|P(\mathbb{N})|=2^{\aleph_{0}}=|\mathbb{R}|$ - this is the original form of the Q.

It's written terribly, and that's a part of why I failed to prove it. the Q is essentially: prove that the cardinality of the SET of all the infinite sub-sets of $\mathbb{N}$ (referred as $\mathrm{X}$), is equal to $2^{\aleph_{0}}$. further more, I'd rather prove that without using onto/one-to-one function, only if possible, in the tools of elementary set theory (meaning, without ZFC)

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An idea:

Prove first that the set of all finite subsets of the naturals is countable:

$$\;\left|\left\{\,X\subset\Bbb N\;/\;|X|<\aleph_0\,\right\}\right|=\aleph_0$$

and then use directly that $\;|P(\Bbb N)|=2^{\aleph_0}\;$