boundedness theorem question Use the Boundedness theorem to show that if the function $f:[0,1]\rightarrow\mathbb{R}$ is continuous and $f(x)$ is not equal to $0$ for all $x ∈ [0,1]$, then there exists $\delta>0$ such that $|f(x)|>\delta$ for all $x ∈ [0,1]$.
I don't understand what to do here.
 A: Someone please edit it if it is wrong in any way or it needs more justification!!
First, let's notice that either $ f(x) > 0 $ or $ f(x) < 0 $ for all $ x \in [0,1] $ (otherwise, since $ f $ is continuous, we can use Darboux property to show that $ f(x) = 0 $ for some $ x \in [0,1]$). We can therefore assume:
$\textbf{Case 1}$: $ f(x) > 0 $. Now, by the mentioned theorem, by continuity we know that $ f $ achieves its extreme points and since $ \forall_{x\in[0,1]} f(x) \neq 0 $, there exists $\delta_1>0$ $ \delta_1 = \inf\limits_{x \in [0,1]}|f(x)| = f(x) > 0 $ and $ f(x) \geq \delta_1 > 0 $ In this case $ \delta = \delta_1 $ 
$\textbf{Case 2}$: Now assume $f(x)<0$. By the boundedness theorem and by continuity, we know that f achieves its extreme points and since $f(x)$ is not equal to $0$ for all $x∈[0,1]$, there exists $\delta_2<0$ such that $\delta_2=\sup f(x)<0$ and $f(x)\leq\sup f(x)<0$. In this case $ |f(x)|= - f(x) >  \delta = -\delta_2> 0 $
A: First, let's notice that either $ f(x) > 0 $ or $ f(x) < 0 $ for all $ x \in [0,1] $ (otherwise, since $ f $ is continuous, we can use Darboux property to show that $ f(x) = 0 $ for some $ x \in [0,1]$). We can therefore assume $ f(x) > 0 $. Now, by the mentioned theorem, by continuity we know that $ f $ achieves its extreme points and since $ \forall_{x\in[0,1]} f(x) \neq 0 $ we have $ \delta = \inf\limits_{x \in [0,1]} f(x) > 0 $ and $ f(x) \geq \delta > 0 $ 
