Our definition of Borel set is an element of the intersection of all $\sigma$-algebras that contains all open sets. Now, I redefined the Borel sets as follows: (i) every open set is a Borel set; (ii) a complement of a Borel set is a Borel set; and (iii) a countable union of collection of Borel sets is a Borel set.
To prove that $f^{-1}(B)$ is a Borel set whenever $f$ is Borel measurable and $B$ is Borel set, I used the inductive approach. First, I proved that the proposition holds for every open set. Second, I proved that if $B$ is a Borel set for which the proposition holds, then it also holds for $\widetilde{B}$. Third, if $\{B_{n}\}_{n=1}^{\infty}$ is a collection of Borel sets for which the proposition is true, then it is also true for $\bigcup_{n=1}^{\infty}{B_{n}}$. Is this approach valid?
Yes, this approach should be right. Your result suffices as the definition of a Borel measurable function, so what definition are you starting with? If $f$ is real-valued, I have often seen the definition that $f$ is Borel measurable if for all $a \in \mathbb{R}$, the preimage $f^{-1}\left( (a,\infty) \right)$ is a Borel set. One can write $(a,\infty)^\mathsf{C} = (-\infty,a]$ (using $(a,\infty)^\mathsf{C}$ to denote the complement), and one can establish that $f^{-1}\left( (-\infty,a] \right) = \left( f^{-1}\big( (a,\infty) \big) \right)^\mathsf{C}$, so $f^{-1}\left( (-\infty,a] \right)$ is Borel measurable. If $a < b$, then $f^{-1}\left( (a,b] \right) = f^{-1}\left( (a,\infty) \cap (-\infty,b] \right) = f^{-1}\left( (a,\infty) \right) \cap f^{-1}\left( (-\infty,b] \right)$, so the half-open intervals $(a,b]$ have Borel measurable preimages. Since the open interval $(a,b)$ can be written as the countable union of intervals $(a,b_n]$ where $b_n \to b^-$, you can show that $f^{-1}\left( (a,b) \right)$ is Borel measurable, and since the open sets in $\mathbb{R}$ are countable unions of intervals $(a,b)$, you can proceed by taking countable unions and complements to show that open sets have Borel measurable preimages, and then that all Borel sets do as well. Does your approach look something like this? What you described is definitely the right idea.