If $X \subset \mathbb{R}$ and $f:X \rightarrow \mathbb{R}$ the $X$is a Borel set because $X = f^{-1}(\mathbb{R})$

Question

In Reference 1 (p. 33), Axler writes the statement: If $X \subset \mathbb{R}$ and there exists a Borel measurable function $f: X \rightarrow \mathbb{R}$, the $X$ must be Borel set [because $X = f^{-1}(\mathbb{R})$]. We suppose that $(X,\mathcal{S})$ is a measurable space. Below is my attempt to prove that statement. In particular, to show that $X = f^{-1}(\mathbb{R})$.

By Assumption, $f$ is a Borel measurable function. Then using the Theorem on ``Condition for measurable function'' Reference 1 (p. 32), we can write that $f^{-1}((a,\infty)) \in \mathcal{S}$ $\forall a \in \mathbb{R}$.

The condition is for all $a \in \mathbb{R}$, then let $a = k \in \mathbb{N}$ and we can write: \begin{equation} \cup_{k=1}^{\infty} (-k,\infty) = (-\infty,\infty) = \mathbb{R}. \end{equation} For each $k \in \mathbb{N}$, $f^{-1}((-k,\infty)) \in \mathcal{S}$, using algebra of inverse image, we can write: \begin{equation} f^{-1}(\cup_{k=1}^{\infty} (-k,\infty))= \cup_{k=1}^{\infty} f^{-1}(-k,\infty)) = f^{-1}(\mathbb{R}) \in \mathcal{S} \end{equation} and by Assumption, we obtain that $f^{-1}(\mathbb{R})$ is a Borel set.

Consider $f^{-1}(\mathbb{R} \setminus \cup_{k=1}^{\infty} (-k,\infty))$. Using algebra of inverse image we obtain: \begin{equation} f^{-1}(\mathbb{R} \setminus \cup_{k=1}^{\infty} (-k,\infty)) = X \setminus f^{-1}(\cup_{k=1}^{\infty} (-k,\infty)) \end{equation} but $\mathbb{R} \setminus \cup_{k=1}^{\infty} (-k,\infty) = \emptyset$

Hence, with $f^{-1}(\emptyset) = \emptyset$, the above equation becomes \begin{equation} f^{-1}(\emptyset) = \emptyset = X \setminus f^{-1}(\cup_{k=1}^{\infty} (-k,\infty)) \end{equation}

By the definition of a complement of any set we have that \begin{equation} X = f^{-1}(\cup_{k=1}^{\infty} (-k,\infty)) \end{equation} and we obtain: \begin{equation} X = f^{-1}(\mathbb{R}) \end{equation}

Since $f^{-1}(\mathbb{R}) \in \mathcal{S}$ is a Borel set, then $X \in \mathcal{S}$ and is also a Borel set. This completes the proof.

Hoping this question is appropriate for this forum, I would like to know if the above proof is correct even if it is a bit long.

Reference: 1 - Axler, S., ``Measure, Integration & Real Analysis'', Springer Open, Cham, 2020.