Let $\epsilon\gt 0$. You know that there exists $\delta_0$ such that any two points that are $\delta_0$ close and both are in $(a,c)$ will have images that are $\epsilon$-close. And you know that there is a $\delta_1$ such that any two points that are $\delta_1$ close and both are in $(c,d)$ will have images that are $\epsilon$-close. Taking the minimum of $\delta_1$ and $\delta_2$ will guarantee any two points that are in the same subinterval and are within $\delta$ of each other will have images that are $\epsilon$-close. The problem lies with the situation in which one point is in $(a,c)$ and the other point is in $(c,b)$.
Now, we know the function is continuous at $c$, so we also know there is a $\delta_2$ such that if a point is within $\delta_2$ of $c$, then its image is within $\epsilon$ of the image of $c$.
Now think about the triangle inequality, and try to put it all together...
(and don't repost the question yet again!)