Continuous Function Greater than 0 Let $f$ be continuous on $[a, b]$. Suppose $f(x) > 0$ for all $x \in [a, b]$. I'm trying to show that there exists a $\alpha > 0$ such that $f(x) > \alpha$ for all $x \in [a, b]$.
I tried to prove this by contradiction. 
Assume that for every $\alpha > 0$, there exists an $x \in [a, b]$ such that $f(x) \leq \alpha$. Then I let $\alpha_n = \frac{1}{n} > 0$. Then there exists an $x_n \in [a, b]$ such that $f(x_n) \leq \alpha_n$. But note that $\alpha_n \to 0$ as $n \to \infty$. This implies that there is an $x_n$ such that $f(x_n) \leq 0$, which is a contradiction. 
Could someone give me feedback on my proof?

It's not correct. Where do you get this $x_n$?  
But, by the compactness of $[a,b]$, your argument can be salvaged. There is a subsequence of $(x_n)$ that converges to an $x\in[a,b]$; and by the continuity of $f$, we would have $f(x)\le0$.
Or, arguing directly, you could consider the minimum value of $f$ on $[a,b]$ (which exists, since $[a,b]$ is closed and bounded).
