Show that continuous functions on $\mathbb R$ are Borel-measurable

You need to establish the following lemma:

Let $(X,\Sigma),(Y,\mathrm{T})$ be measurable spaces. Suppose that $\mathrm{T}$ is generated by $\mathcal{E}$. A mapping $f:X \to Y$ is $(\Sigma,\mathrm{T})$-measurable if and only if $$\forall E \in \mathcal{E}, f^{-1}[E] \in \Sigma$$

Proof: The only if part is trivial. For the other direction, observe that the $\sigma$-algebra generated by the set $\{S \subseteq Y\mid f^{-1}[S] \in \Sigma\}$ contains $\mathcal{E}$, so it also contains $\mathrm{T}$.

By substituting $\Sigma,\mathrm{T}$ with the Borel sets, and $\mathcal{E}$ with the open sets (a.k.a. topology) on $\mathbb{R}$, the proof is complete.

This relies basically on the fact that inverse functions respect all three set operations: arbitrary unions, arbitrary intersections and complements. In other words $f^{-1}(\cup B_\alpha)=\cup f^{-1}(B_\alpha)$, $f^{-1}(\cap B_\alpha)=\cap f^{-1}(B_\alpha)$ and $f^{-1}(A^c)=(f^{-1}(A))^c$. All three of these are pretty easy to prove. Incidentally functions going forwards only respect unions, while inverse is much better behaved.