If $f:[0,1] \rightarrow \mathbb{R}$ is continuous then it is uniformly continuous I do have a general proof for this problem. That is if there is a continuous function on a compact interval then the function is uniformly continuous.
But I'm wondering whether there is a different (more simpler one) for this particular domain. 
 A: Wether there is a different or a simpler proof is not the same question. You can find proof for any closed interval of $\mathbb{R}$ that do not explicitly call compactness ( an example is to be found in Michael Spivak's textbook Calculus, as @MathematicsStudent1122 stated in his comment, see page 142 of https://notendur.hi.is/mbh6/html/_downloads/Calculus%20-%20Michael%20Spivak.pdf). But all these proofs are based on the ability to explicitly exhibit a finite open cover for intervals of $\mathbb{R}$, which in the end, is just as strong as calling compactness. 
If f is differentiable, then the mean-value theorem will definitely give you a shortcut, but once again, the ability to find an upper bound for the derivative will be based on the compactness of $[0,1]$, wether explicitly or implicitly. 
As to know if the above proofs will be simpler, there are some cases in mathematics (more frequent than you would think) where getting more general makes simpler proof. I reckon this is one of these cases. What we can say however, is that not calling explicitly compactness is gonna use less advanced knowledge with regard with a basic cursus in mathematics. 
