Banach fixed point theorem, $T: D \to V, f \to Tf$

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I am working on a problem, but i am not sure if my solutions are correct.

To the Problem:

Let $ (V,‖·‖) = (C([0,1]),‖·‖∞)$ be a normalized vector space. and let $D:=\{f∈V; \| f \| \leq 1/3\}$

We define T :

$T:D→V$ , $f→Tf$ , $(Tf)(x) =∫_0^x\frac {s^2}{2}+f(s)^2ds$ $(x∈[0,1]).$

(a) Show that for all $f_1$ and $f_2$: $$‖Tf_1−Tf_2‖≤\frac {2}{3}·‖f_1−f_2‖.$$

for (a) i have :

$$T' = Tf' = \frac{x^2}{2}+f(x)^2 $$ and then $$0 < Tf'(0) < Tf'(1) = \frac{11}{18} \le \frac{2}{3}$$

After that i am gonna show with mean value theorem it satisfies : $$‖Tf_1−Tf_2‖≤\frac {2}{3}·‖f_1−f_2‖$$

Is my solution for (b) acceptable? If no, how could i improve my solution?