Continuous function and Borel sets If $f:\mathbb{R}^p \rightarrow \mathbb{R}$ is continuous and $B\subset \mathbb{R}$ is Borel set, how to show that $f^{-1}(B)$ is also Borel set?

I was trying to construct a $\sigma$-ring of sets whose preimage is Borel, but they're not necessarily all borel in that sigma ring...

What would be the main idea for the proof here?
Hint: prove that $\mathcal A := \{B \in B(\Bbb R): f^{-1}(B) \in B(\Bbb R^p)\}$ is a sigma algebra containing the open sets, where $B(\Bbb R^d)$ is the Borel sigma algebra on $\Bbb R^d$.

Added:

This way, $\tau \subset \mathcal A$ (where $\tau$ is the usual topology of $\Bbb R$), "applying $\sigma$ both sides" we get $\sigma(\tau)\subset \cal A$, therefore $B(\Bbb R)\subset \cal A$. Which means that each $B \in B(\Bbb R)$ is in $\cal A$, i.e. $f^{-1}(B) \in B(\Bbb R^p)$.
A Borel set is any set in a topological space that can be formed from open sets through the operations of countable union, countable intersection, and relative complement.

As the inverse images of open sets under a continuous function are open sets and inverse images of a countable union is the countable union of the inverse images, same for countable intersection and relative complement. We get the the inverse image of a Borel set under a continuous function is a Borel set.