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: 
I didn't want to answer (just to comment), but since some other guy answered what I commented, then let me write a simpler answer.

The identity function you propose always works. Indeed, if your $\sigma$-algebra does not contain all Borel subsets then it does not contain at least an open subset of the reals. Then just take the inverse image of that open set through the identity function.
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.