Prove the following: Let $\displaystyle a < b $ be real numbers. and let $\displaystyle f: [a,b] \to \bold{R} $ be a function which is continuous on $\displaystyle [a,b] $. Then $\displaystyle f $ is also uniformly continuous.

Assume for contradiction that $\displaystyle f $ is not uniformly continuous. Then there are two equivalent sequences $\displaystyle (x_{n})_{n=0}^{\infty} $ and $\displaystyle (y_{n})_{n=0}^{\infty} $ in $\displaystyle [a,b] $ such that the sequences $\displaystyle (f(x_{n}))_{n=0}^{\infty} $ and $\displaystyle (f(y_{n}))_{n=0}^{\infty} $ are not equivalent. Then we let $\displaystyle E = \{n \in \bold{N}: f(x_{n}) \ \text{and} \ f(y_{n}) \ \text{are not} \ \varepsilon \ \text{close} \} $. Then use Bolzano-Weierstrass Theorem to obtain contradiction?

Is this correct?