Answer
There is an interval $(c-\delta,c+\delta)$ around $c$ such that $f$ has the same sign as $f(c)$. The proof is discussed in detail in the Work Step by Step.
Work Step by Step
We have $f$ is continuous at $c$.
This means according to definition, $\lim_{x\to c}f(x)=f(c)\ne0$
According to the formal definition of limit, since $\lim_{x\to c}f(x)=f(c)$, for any values of $\epsilon\gt0$, there exists a corresponding value of $\delta\gt0$ such that for all $x$ $$0\lt|x-c|\lt\delta\Rightarrow |f(x)-f(c)|\lt\epsilon$$
Consider the inequality: $$|f(x)-f(c)|\lt\epsilon$$ $$-\epsilon\lt f(x)-f(c)\lt\epsilon$$ $$-f(x)-\epsilon\lt-f(c)\lt-f(x)+\epsilon$$ $$f(x)+\epsilon\gt f(c)\gt f(x)-\epsilon$$
Now if we take $\epsilon=f(x)/4$, then $$\frac{5}{4}f(x)\gt f(c)\gt \frac{3}{4}f(x)$$
Since $f(c)\in(3/4f(x),5/4f(x))$, $f(c)$ must have the same sign as $f(x)$, in the mentioned interval $$0\lt|x-c|\lt\delta$$ $$-\delta\lt x-c\lt\delta$$ $$c-\delta\lt x\lt c+\delta$$
Hence, there is an interval $(c-\delta,c+\delta)$ around $c$ such that $f$ has the same sign as $f(c)$.
