Proof without using Extreme value theorem? $$\text{Let } f:[a,b]\to \mathbb{R}  \text{ be continuous and strictly positive. } $$
$$ \text{prove } c= \inf f(\,(a,b)\,)\neq 0$$
Is there a way other than using the extreme value theorem?
One way might be to show $c= \min f( [a,b] )$, and therefore positive, but I'm not sure how to proceed (without the theorem).
If $ c=0 $ intuition tells me there must be a point $x_0 \in [a,b]$ such that $\lim_{x \to x_0} f(x) = 0 $, which contradicts its continuous and positive... 

Suppose that $\inf\{f(x):x\in(a,b)\}=0$. Then, for each positive integer $n$, there is some $x_n\in(a,b)$ such that $$(*)\quad 0<f(x_n)<\frac{1}{n}.$$ Since $[a,b]$ is compact and is the closure of $(a,b)$, the sequence $(x_n)$ has a subsequence $(x_{n_k})$ that converges to some $x^*\in[a,b]$. Since $f$ is continuous, $$\lim_{k\to\infty}f(x_{n_k})=f(x^*).$$ But this limit must be zero because of $(*)$, implying that $f(x^*)=0$, which is a contradiction.
