Solve $||T||$ and show that with mapping $I:C([0,1],\mathbb{R})\to C([0,1],\mathbb{R})$ where Ix=x we get that $||I+T||=1+||T||.$
I have managed to find the way of equation: $||T||=\sup_{f\neq 0}\frac{||Tf||}{||f||}=\sup_{f\neq 0}\frac{\sup_{t\in [0,1]}|Tf(t)|}{\sup_{t\in [0,1]}|f(t)|}$, but don't know how to continue.
I have also managed to show: $||I+T||\leq ||I||+||T||=\sup_{f\neq 0}\frac{||If||}{||f||}+||T|| = \sup_{f\neq 0}\frac{||f||}{||f||}+||T||=1+||T||$, but that's only the difference and not the whole solution.
Since $T$ is linear, note that $||T||=\sup_{||f|| = 1}||Tf(t)||$.
Now, for $f$ to have norm $1$ means that its largest value is 1 (in absolute value). Now consider $Tf(t) = t^2f(t)$. Under the above constraint, it is clear that this is less than $1$. Therefore this gives an upper bound on the norm of $T$.
To see that this is actually the norm of $T$, consider the function $f(t) = 1$. Clearly this has norm $1$, and $Tf(1) = 1$, which gives you the lower bound of 1.