Continuity on finite interval

Question

Problem: $f$ is a continuous function on $[a,b]$, and $f(x)>0$ for all $x$ in $[a,b]$. Prove that there is an $\alpha>0$ such that $f(x)>\alpha$ for all $x$.

Please help in the steps of this proof!

Answer 1

If not, $\exists~(x_n)\subset[a,b]$ such that $f(x_n)\le\dfrac{1}{n}~\forall~n\in\mathbb Z^+.$

$(x_n)$ being bounded it has a convergent subsequence $(x_{r_n})\subset[a,b].$

Suppose $x_{r_n}\to c$ in $[a,b].$ Due to the continuity of $f$ at $c,~f(x_{r_n})\to f(c).$

But $0<f(x_{r_n})\le\dfrac{1}{r_n}\le\dfrac{1}{n}\to0\implies f(x_{r_n})\to0\implies f(c)=0!$