Prob. 2, Chap. 6, in Baby Rudin: If $f\geq 0$ and continuous on $$ with $\int_a^bf(x)\ \mathrm{d}x=0$, then $f=0$

Here is Prob. 2, Chap. 6, in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

Suppose $f \geq 0$, $f$ is continuous on $[a, b]$, and $\int_a^b f(x) \ \mathrm{d} x = 0$. Prove that $f(x) = 0$ for all $x \in [a, b]$.

My Attempt:

Let $c$ and $d$ be any points in $[a, b]$ such that $c \leq d$. As $f(x) \geq 0$ for all $x \in [a, b]$ and as $f$ is continuous on $[a, b]$, so $f(x) \geq 0$ for all $x \in [c, d]$ and $f$ is continuous on $[c, d]$; therefore $f \in \mathscr{R}$ on $[c, d]$ and
$$ \int_c^d f(x) \ \mathrm{d} x \geq \int_c^d \hat{0}(x) \ \mathrm{d} x = 0$$
by Theorem 6.12 (b) in Baby Rudin, where $\hat{0}$ denotes the zero function on $[c, d]$. That is,
$$ \int_c^d f(x) \ \mathrm{d} x \geq 0 \tag{0}$$
for any points $c$ and $d$ such that $a \leq c \leq d \leq b$.

Suppose that there is a point $p \in [a, b]$ such that $f(p) > 0$. Then, as $f$ is continuous at $p$, so, for any real number $\varepsilon$ such that
$$ 0 < \varepsilon < \frac{ f(p) }{2}, $$
we can find a real number $\delta > 0$ such that
$$ \lvert f(x) – f(p) \rvert < \varepsilon < \frac{ f(p) }{2} $$
for all $x \in [a, b]$ for which
$ \lvert x-p \rvert < \delta$.

Therefore, we can conclude that
$$ \frac{ f(p) }{2} < f(x) < \frac{3 f(p) }{2} \tag{1} $$
for all $x \in I$, where
$$I \colon= [a, b] \cap \left[ p- \frac{\delta}{2}, p+ \frac{\delta}{2} \right]. $$
Let’s put $I \colon= [u, v]$. Then we see that
$$
\begin{align}
\int_a^b f(x) \ \mathrm{d} x &= \int_a^u f(x) \ \mathrm{d} x + \int_u^v f(x) \ \mathrm{d} x + \int_v^b f(x) \ \mathrm{d} x \\
& \qquad \qquad \mbox{ [ by an extension of Theorem 6.12 (c) in Baby Rudin ] } \\
&\geq \int_u^v f(x) \ \mathrm{d} x \qquad \mbox{ [ by (0) above ] } \\
&\geq \int_u^v \frac{f(p)}{2} \ \mathrm{d} x \qquad \mbox{ [ using (1) and Theorem 6.12 (b) in Baby Rudin ] } \\
&= \frac{f(p)}{2} (v-u) \\
&> 0, \qquad \mbox{ [ by (2) and our choice of $u$ and $v$ above ] }
\end{align}
$$
which contradicts our hypothesis that $\int_a^b f(x) \ \mathrm{d} x = 0$.

Is this proof sound enough in terms of its logic and rigor? If so, then is it also lucid enough in its presentation?

Solutions Collecting From Web of "Prob. 2, Chap. 6, in Baby Rudin: If $f\geq 0$ and continuous on $$ with $\int_a^bf(x)\ \mathrm{d}x=0$, then $f=0$"