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?

For otherwise $f$ becomes arbitrarily close to $0.$ Consequently $\exists$ a sequence $\{x_n\}$ in $[a,b]$ such that $$0<f(x_n)<\frac{1}{n}$$
Squeezing $f(x_n)\to 0.$ 
Since $\{x_n\}$ is bounded $\exists$ a convergent subsequence $\{x_{r_n}\}$ of $\{x_n\}$ converging to some $l\in[a,b].$
$f$ is continuous at $l$ and $x_{r_n}\to l\implies f(x_{r_n})\to f(l)\neq 0!$
