A bijective function $f:\Bbb R\to [0, \infty)$ has infinitely many discontinuities.

Question

A bijective function $f:\Bbb R\to [0, \infty)$ has infinitely many discontinuities.

Suppose that $f$ has finitely many discontinuities say $\{x_1,x_2,\ldots ,x_n\}$.Then $f$ is continuous on $(-\infty,x_1)\cup (x_1,x_2)\cup \ldots \cup (x_{n-1},x_n)\cup (x_n,\infty)$.

So $f$ is monotone on each of the sub-intervals.

Obviously $f(x_i,x_j)\cap f(x_j,x_k)=\emptyset$ since $f$ is bijective.

Now $f(x_i,x_j)$ will be an interval since $f$ is injective.Also it will be an open interval since $f$ is continuous.

Now $[0,\infty)\setminus\cup \{(f(x_i,x_j))\}\cup f(-\infty ,x_1)\cup f(x_n,\infty)$ must have $n+1$ points since $f(x)\neq 0\forall x$ and $f(x)\neq f(x_1),f(x_2)\ldots f(x_n)\forall x$

But $\Bbb R\setminus \{x_i\}$ has $ n$ points.

So we have a contradiction.

Is the proof correct?Please suggest required edits