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I do not know how to prove or how to even start to proving that statement: $$|N|<|R|$$

Thanks for all your answers.

Niki Sedlarevič
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    You might not know how a finished proof starts, but I think you know some relevant things. For instance, what does it mean that one cardinality is strictly greater than another? That's a big (and frankly rather obvious) clue as to where you should begin looking for a proof. And if you don't know what it means, then again you have a big clue as to what you should do first: find out what it means. – Arthur Jul 18 '19 at 08:57

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Well, the power set $P(X)$ of a set $X$ has always higher cardinality.

Indeed, the power set $P(X)$ cannot have smaller cardinality than $X$, since for each $x\in X$, $\{x\}$ lies in $P(X)$.

Suppose there is a bijection $f:X\rightarrow P(X)$. Define the sets $A = \{x\in X\mid x\in f(x)\}$ and $B=\{x\in X\mid x\not\in f(x)\}$.

Since $f$ is a bijection and $B\subseteq X$, there is $b\in X$ with $f(b)=B$.

Now, $b\in B$ implies $b\not\in f(b)$ implies $b\not\in B$, and $b\not\in B$ implies $b\in f(b)$ implies $b\in B$. A contradiction.

Wuestenfux
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This was proved by Cantor, with his diagonal argument. He showed the power set of $X$ has higher cardinality than $X$, for any set $X$. That is $\vert P(X)\vert\gt \vert X\vert$.

But, it's easy to show $\vert P(\Bbb N)\vert=\vert \Bbb R\vert$. See my question here.