elementary set theory - Proof that a subset of R has same cardinality as R

I am sitting with a task where I have to prove the following:

Claim:
Every subset of $\mathbb{R}$, that contains an interval $I$ with $a < b$, has the same cardinality as $\mathbb{R}$.

So I think that I should prove that there exist a bijection from $I$ to $\mathbb{R}$? I'm kinda lost, and don't know how to start.

There is a lemma 3 in the book saying:
Let $a,b \in \mathbb{R}$ with $-\infty\neq a < b \neq \infty$. There exist a $f\colon ]-1; 1[ \to ]a; b[$ that is bijektiv. The intervals $]-1; 1[$ and $]a; b[$ has same cardinality.

Another lemma 4 says:
$f\colon ]0; 1[ \to ]1; \infty[$, $x\to \frac{1}{x}$, is bijective. The intervals $f\colon ]0; 1[ \to ]1; \infty[$ have same cardinality.

(There is an image added to lemma4)

The above interval $]-1; 1[$ has same cardinality as $\mathbb{R}$, and the $f$ is bijective
Any help is highly appreciated

Answer

Hints:

First prove that any two non-empty open intervals have the same cardinality.

Second, pass now to use the nice interval $\;(-\pi/2\,,\,\pi/2)\;$ and a rather nice, simple trigonometric function to show equipotency with $\;\Bbb R\;$