Since every $x \in C[a, b]$ is continuous on a compact set, it's bounded.
For any $x \in C[a, b]$, we have:
$$
|f(x)| = \left|\int_a^b x(t) x_0(t) \, dt \right| \le \int_a^b |x(t)| \cdot |x_0(t)| \, dt \le \|x\| \int_a^b |x_0(t)| \, dt
$$
Thus, $f$ is bounded and its norm satisfies:
$$
\|f\| \le \int_a^b |x_0(t)| \, dt \newcommand{sgn}{\operatorname{sgn}}
$$
In fact, equality holds. To see this, consider the sign function $\hat x(t) = \sgn(x_0(t))$. By Lusin's theorem, there is exists a sequence of functions $x_n \in C[a, b]$ such that $\|x_n\| \le 1$ and $x_n(t) \to \hat x(t)$ as $n \to \infty$ for every $t \in [a, b]$. By the dominated convergence theorem, we have:
\begin{align}
\lim_{n \to \infty} f(x_n) &= \lim_{n \to \infty} \int_a^b x_n(t) x_0(t) \, dt \\
&= \int_a^b \lim_{n \to \infty} x_n(t) x_0(t) \, dt \\
&= \int_a^b \sgn(x_0(t)) x_0(t) \, dt \\
&= \int_a^b |x_0(t)| \, dt
\end{align}
For the second linear functional you added, approximate the following function via Lusin's theorem in a similar manner:
$$
\hat x(t) = \begin{cases} 1 & \text{if } t \le \frac{a+b}{2} \\
-1 & \text{if } t > \frac{a+b}{2}
\end{cases}
$$
Hint: It is an easy exercise to prove that $$\|f\|=\sup_{\|\phi\|\leq1}|f(\varphi)|=\inf\{c>0: |f(\varphi)|\leq c\|\varphi\| \text{ for all }\varphi \}.$$
Since you have shown that $|f(\varphi)|\leq\|\varphi\|$ for all $\varphi$, this shows that $\|f\|$, which is the least constant with this ability, is less than $1$, which (as you showed) has this property.
Best Answer
Prove by contradiction. Suppose $|\phi (f)|=1$. Then $1=|\int_0^{1/2}f-\int_{1/2}^{1}f| \leq \frac 1 2 +\frac 1 2 =1$. Hence we must have $|f(x)|=1$ for all $x$. [Even if $|f(x)|<1$ at one point we get strict inequality and we get $1<1$!]. Thus $f$ can take only the values $1$ and $-1$. But $f$ is continuous, so either $f \equiv 1$ or $f \equiv -1$. But then $\phi (f)=0$. [I am assuming that you are considering real valued continuous functions but the result is true in the complex case also].