I am working on the Exercise $1.17$ here.These are my attempt as followings:
It is clear that there exists a non-zero function $f\in E^*$. From equation $2$, we know that $\text{dist}(u,M)=\frac{|<f,u>|}{\left \| f \right \| }$. According to $1.3$, we know that $\||f|| = 1$, which implies that $\text{dist}(u,M) = |\int_{0}^{1}u(t)\,\mathrm{d}t|$. Suppose the infimum can be reached for some $u_0$, i.e., for such $u_0 \in E-M$, there exists a corresponding $v_0 \in M$ such that $$\inf_{v\in M}\left \| u_0-v \right \|=\left \|u_0-v_0 \right \|. $$
And $$\text{dist}(u_0,M)=|\int_{0}^{1}u(t)\,\mathrm{d}t|, \text{i.e.},|\int_{0}^{1}u_0(t)-v_0(t)\,\mathrm{d}t|=\left \| u_0-v_0 \right \| =\max_{t\in[0,1]}|u_0(t)-v_0(t)|.$$ Since $u_0, v_0 \in E$, we have $(u_0 - v_0)(0) = 0$, which is obviously impossible.
Next, we prove: Since $u_0(t),v_0(t)$is continuous, suppose $$\left \| u_0(t)-v_0(t)\right \| =(u_0-v_0)(t_0),$$ $\exists t_\delta\in(0,t_0) $ such that $$\ (u_0-v_0)(t)\leq \frac{1}{2}\left \| u_0(t)-v_0(t)\right \|,\forall t\in[0,t_\delta)$$
Then \begin{align} \int_{0}^{1}u_0(t)-v_0(t)\,\mathrm{d}t&=\int_{0}^{t_\delta}u_0(t)-v_0(t)\,\mathrm{d}t+\int_{t_\delta}^{1}u_0(t)-v_0(t)\,\mathrm{d}t\\ &\leq(1-\frac{1}{2}t_\delta)\left \| u_0(t)-v_0(t)\right \|<\left \| u_0(t)-v_0(t)\right \| \end{align} it contradicts.