Continuity Properties 
Let $f$ be continuous on $[a,b]$, and suppose that $f(c)\not=0$ for some $c\in(a,b)$. Show that there exists $\delta>0$ with the property that $\forall x\in (c-\delta,c+\delta):f(x)\not=0$. 

Proof: $f$ is continuous so $\forall x\in(c-\delta,c+\delta):|f(x)-f(c)|<\epsilon \implies -\epsilon <f(x)-f(c)<\epsilon \implies f(c)-\epsilon <f(x)<f(c)+\epsilon$ And here I get stuck. 
How can i prove this proposition?
 A: Since $f(c)\neq 0$, there exists $\varepsilon>0$ so that $B(f(c),\varepsilon)\cap B(0,\varepsilon)=\emptyset$. Since $f$ is continuous, there exists $\delta>0$ so that $fB(c,\delta)\subseteq B(f(c),\varepsilon)$, and we may assume that $B(c,\delta)=(c-\delta,c+\delta)$ by choosing smaller $\delta>0$ if necessary.
Now, what can you say about $f(x)$ for $x\in (c-\delta,c+\delta)$?
A: The contrapositive of 
$$
\exists \delta >0\,\forall x\in (c-\delta,c+\delta):\, f(x)\neq 0
$$
is
$$
\forall \delta >0\,\exists x\in (c-\delta,c+\delta):\, f(x)=0.
$$
Assume that the contrapositive holds and try to reach a contradiction (contradicting that $f$ is continuous at $c$ and $f(c)\neq 0$).
A: If you can construct an $\varepsilon$ neighborhood around $f(c)$ which discludes zero, then since $f$ is continuous you can find a $\delta > 0$ where if $x \in [a,b]$ and $\left\lvert x - c \right\rvert < \delta$ then $\left\lvert f(x)-f(c) \right\rvert < \varepsilon$.
A: $\forall x\in(c-\delta,c+\delta)/\{c\}:||f(x)|-|f(c)||<|f(x)-f(c)|<\epsilon$
take $\epsilon= \frac{|f(c)|}{2}$,then$|f(x)|>\frac{|f(c)|}{2}$.
since $f(c)\neq 0$,therefore $f(x)\neq 0$
