Let A,B,C be nonempty sets of real numbers, and let f:A->B and g:B->C be uniformly continuous functions. Prove that the composition gof:A->C is also unfiormly continuous,

Here's what I have. Is it correct?

Let$\displaystyle \epsilon >0$ then there is $\displaystyle \delta>0$ s.t for all x1,y1 $\displaystyle \in A$ if $\displaystyle |x_1-y_1|<\delta$ then |$\displaystyle f(x_1)-f(y_1)|<\epsilon$ so f(x1),f(y1)$\displaystyle \in$B. Now let $\displaystyle \epsilon$>0 then there exists $\displaystyle \delta$>0 s.t for all x2,y2$\displaystyle \in$B if |x2-y2|<$\displaystyle \delta$ then |g(x2)-g(y2)|<$\displaystyle \epsilon$. So |(gof)(x1)-(gof)(y1)|=|g(f(x1))-g(f(y1))|<$\displaystyle \epsilon$.