Let $0leq a leq b leq 1$. Then we have for all natural numbers $mgeq 2$ the inequality $b^{frac m2}-a^{frac m2} leqfrac m2(b-a)$

Other way is: if $x=sqrt{b}, y=sqrt{a}$, then: $$b^{tfrac{m}{2}}-a^{tfrac{m}{2}}=(x-y)sum_{i=1}^{m}x^{m-i}y^{i-1}$$ How $0leq yleq xleq1$ then: $$b^{tfrac{m}{2}}-a^{tfrac{m}{2}}=(x-y)sum_{i=1}^{m}x^{m-i}y^{i-1}leq(x-y)frac{mx+my}{2}$$

There is also the following reasoning.

Let $f(b)=frac{m}{2}(b-a)-b^{frac{m}{2}}+a^{frac{m}{2}}.$

Thus, $$f'(b)=frac{m}{2}-frac{m}{2}cdot b^{frac{m}{2}-1}=frac{m}{2}left(1-b^{frac{m-2}{2}}right)geq0.$$ Id est, $$f(b)geq f(a)=0$$ and we are done!

Your proof is correct but you had to say that $$sup_{[a,b]}fle sup_{[0,1]}f$$

hint

here is an other :

Let $$g(x)=x^{frac m2}$$ $ g $ is continuous at $ [a,b] $ and differentiable at $ (a,b) $, then by MVT,

$$exists cin (a,b);; :;frac{g(b)-g(a)}{b-a}=g'(c)$$ but $$g'(c)=frac m2 c^{frac m2-1}$$ and $$0<c<1$$