Question: Let $\mathcal{A}$ be a σ-algebra on $\mathbb{R}$ that does not contain all Borel sets (i.e that is, it holds
that $\mathcal{B}(\mathbb{R}) \setminus \mathcal{A} \not= ∅)$. Prove that there is a continuous function f : $\mathbb{R} → \mathbb{R}$ which
is not $\mathcal{A}$-measurable

I have a really simple example but I am not sure if it correct. Take $\mathcal{A} = (\emptyset , \mathbb{R} , (0,1) , \mathbb{R}\setminus (0,1))$ This is clearly a sigma algebra.
Then take $f(x)=x$ we have that $f^{-1}((5,6))=(5,6)\notin\mathcal{A}$ so $f$ is not $\mathcal{A}$ measurable and clearly $f$ is continuous.

Is this a correct solution?

**EDIT**

Thanks to some incredibly bright chaps on this site I have seen the error in my solution, I simply misread the question. So for all those who come after here is the answer:

As $\mathcal{B}$, the Borel $\sigma$-algebra, is the smallest $\sigma$ algebra containing all open sets and $\mathcal{B} \not \subseteq \mathcal{A}$ we have at least one interval, $I \in \mathcal{B}$, that is not in $\mathcal{A}$ Hence taking the pre-image of $I$. We have: $\mathcal{B} \ni I, f^{-1}(I) = I \notin \mathcal{A}$ So $f$ is not $\mathcal{A}$ measurable

Answer: 
The way the question is phrased, you are not allowed to choose $\mathcal A$.

But your idea works anyway. If $\mathcal A$ does not contain all Borel sets, then it fails to contain some interval. Now you can do your example using said interval.
Consider a closed set $C\notin \mathcal A$. There is a continuous (or even $C^\infty$) function $f:\Bbb R\to \Bbb R$ such that $C=f^{-1}\{0\}$. Therefore $f$ is not $\mathcal A$-measurable.