Task
1) Show that function $T:C[0, 1] \to C[0, 1]$ is a contraction, then $$T(f)(x)=\int_{0}^x (x-t) f(t) dt,$$ $$x\in [0,1], f\in C[0, 1].$$
2) Find a fixed point of $T(f)(x)$.
My progress
1) I have to show that $$\vert T(f)(x)-T(g)(x)\vert \le \mathcal L \vert f - g \vert,$$ $$\mathcal L \lt 1.$$
Solution:
$\vert T(f)(x)-T(g)(x)\vert = \vert \int_{0}^x (x-t) f(t) dt - \int_{0}^x (x-t) g(t) dt \vert = \vert \int_{0}^x (x-t) (f(t) - g(t)) dt \vert \le \int_{0}^x \vert (x-t) (f(t) - g(t)) \vert dt \le \int_{0}^x \vert x-t\vert dt \int_{0}^x \vert f(t)-g(t)\vert dt = \vert x^2-\frac{x^2}{2} \vert \int_{0}^x \vert f(t)-g(t)\vert dt \le \vert \frac{x^2}{2}\vert \vert f-g\vert$
So $$\vert Tf-Tg \vert \le \vert \frac{x^2}{2}\vert \vert f-g\vert.$$
Then I check if $\vert \frac{x^2}{2}\vert \lt 1.$ If $x\in [0,1]$, the biggest value I get then $x=1$ and it is $\frac{1}{2}$. So $\mathcal L \lt 1.$
Question:
Is the part 1) correct? And what about part 2)?
Your argument is not correct, as the inequality $\int_0^x f_1f_2\leq\int_0^xf_1\int_0^xf_2$ does not hold in general.
As your $T$ is linear, you just need to show that $\|Tf\|\leq c\|f\|$ for some $c<1$.
For that, you have $$ |Tf(x)|\leq\int_0^x(x-t)|f(t)|\,dt\leq\|f\|\,\int_0^x(x-t)dt=\|f\|\,\tfrac{x^2}2\leq\tfrac12\,\|f\|. $$ So $\|Tf\|\leq\tfrac12\,\|f\|$. Your estimate was correct, even if the argument wasn't.
A fixed point for $T$ is an $f\in C[0,1]$ such that $$\tag1f(x)=\int_0^x (x-t) f(t)\,dt=x\int_0^x f(t)\,dt-\int_0^x tf(t)\,dt$$ for all $x\in [0,1]$. As $f$ is continuous, the right-hand-side in $(1)$ is differentiable, so $f$ is differentiable. If we differentiate in $(1)$, we get $$\tag2 f'(x)=\int_0^xf(t)\,dt+xf(x)-xf(x)=\int_0^x f(t)\,dt. $$ Now the right-hand-side in $(2)$ is differentiable, so $f'$ is differentiable. differentiate again, to get $$\tag3 f''(x)=f(x). $$ So $f(x)=c_1\cos x+c_2\sin x$. From $(1)$, we have $f(0)=0$, and from $(2)$ we have $f'(0)=0$. It follows that $f=0$ is the only fixed point of $T$.