Problem
Let $[a,b]$ be a closed and bounded interval, and let $C[a,b]$ be the vector space of all continuous real-valued functions on $[a,b]$. Define the function $\|\cdot\|_1:C[a,b]\to\mathbb{R}$ by letting$$ \|f\|_1=\int_{[a,b]}|f|d\lambda.$$Prove that if a function $f\in C[a,b]$ satisfies $\|f\|_1=0$, then $f=0$.
(Note that $\lambda$ is the Lebesgue measure.)
My Attempt
Let $f\in C[a,b]$. Suppose that $$\|f\|_1=\int_{[a,b]}|f|d\lambda= \sup\left\{\int hd\lambda:h\in\mathscr{S}_+\ \text{and}\ h\leq|f|\chi_{[a,b]}\right\} = 0.$$ Assume to the contrary that $f(x)\neq0$ for some $x\in[a,b]$. Then $|f(x)|>0$. Since $f$ is continuous, $|f|$ must be continuous and there must exists a closed and bounded interval $[c,d]\subseteq[a,b]$ such that $|f(z)|>0$ for all $z\in[c,d]$. Define a function $h$ by letting $h(z)=0$ for all $z\notin[c,d]$ and $h(z)=\min_{z\in[c,d]}|f(z)|$ (note that the minimum exists because $|f|$ is a continuous function on a compact set $[c,d]$). Then $h\in\mathscr{S}_+$ and $h\leq|f|\chi_{[a,b]}$, but $\int hd\lambda>0$, which is a contradiction. Thus, $f=0$.
Some definitions:
Definition$\quad$ Let $X,\mathscr{A}$ be a measurable space. $\mathscr{S}_+$ is the set of simple nonnegative real-valued $\mathscr{A}$-measurable functionson $X$.
Definition$\quad$$\chi_A$ is the characteristic function of the set $A$.
My Question
I am not sure if my attempt is correct. Could someone please help me check? Especially, I claimed that "there must exists a closed and bounded interval $[c,d]\subseteq[a,b]$ such that $|f(z)|>0$ for all $z\in[c,d]$" without proof. (In fact, I am not sure how to prove this rigorously, but I seriously believe this is correct.)