Prove Continuous functions are borel functions Take $f: (a,b) \to \mathbb{R}$ , continuous for all $x_{0}\in (a,b)$ and take $(Ω = (a,b) , F = ( (a,b) ⋂  B(\mathbb{R}))$ where $B(\mathbb{R})$ is the Borel $\sigma$-algebra.
Prove $f$ is a borel function by showing that $\{x \in(a,b): f(x) < c \}$ is in $F$.
I know that continuity of f means that for all $x\in(a,b)$ and all $\varepsilon>0$ there exists a $\delta>0$ such that $|x-x_{0}| < \delta$ implies $|f(x)-f(x_{0})| < \varepsilon$.
But Then I am stuck, how would I use these facts to help me ? 
Thanks in advance for any help

To expand on Thomas E.'s comment: if $f$ is continuous, $f^{-1}(O)$ for $O$ open is again open. 
$\{x \in (a,b) : f(x) < c \} = f^{-1}((- \infty , c)) \cap (a,b)$. Now all you need to show to finish this proof is that $f^{-1}((- \infty , c))$ is in the Borel sigma algebra of $\mathbb R$.
Edit (in response to comment)
Reading your comment I think that your lecturer shows that $S := \{x \in (a,b) : f(x) < c \} $ is open. In a metric space, such as $\mathbb R$ with the Euclidean metric, a set $S$ is open if for all $x_0$ in $S$ you can find a $\delta > 0$ such that $(x_0-\delta, x_0+\delta) \subset S$. 
To show this, your lecturer picks an arbitrary $x_0 \in S$. Then by the definition of $S$ you know that $f(x_0) < c$. This means there exists an $\varepsilon > 0$ such that $f(x_0) + \varepsilon < c$, for $\varepsilon$ small enough. Since $f$ is continuous you know you can find a $\delta_1 > 0$ such that $x \in (x_0 - \delta_1, x_0 + \delta_1) $ implies that $|f(x_0) - f(x)| < \varepsilon$. Now you don't know whether $(x_0 - \delta_1, x_0 + \delta_1) $ is contained in $(a,b)$. But you know that since $(a,b)$ is open you can find a $\delta_2 > 0$ such that $(x_0 - \delta_2, x_0 + \delta_2) \subset (a,b)$. Now picking $\delta := \min (\delta_1, \delta_2)$ gives you that $(x_0 - \delta, x_0 + \delta) \subset S$ because $(f(x_0) - \varepsilon, f(x_0) + \varepsilon) \subset (-\infty , c)$.
