If $\int_0^x f \ dm$ is zero everywhere then $f$ is zero almost everywhere

It is sufficient to prove that $f$ is zero almost everywhere on any bounded interval.

(1) By additivity it is easy to see that $$\int_a^bf(x)dx=\int_0^bf(x)dx - \int_0^af(x)dx$$ for all bounded intervals $(a,b)$ (and also for $[a,b)$, $(a,b]$ and $[a,b]$).

(2) Using (1) it is easy to see that $$\int_Bf(x)dx=0$$ for any bounded Borel measurable set.

(3) Any Lebesgue measurable set $A$ is of the form $A=B\cup Z$ where $B$ is a Borel measurable set and $Z$ is a set of measure zero. Hence, by (2) we acheive $$\int_A f(x)dx= 0$$ for any bounded Lebesgue measurable set $A$.

(4) Now look at the sets $A_+(n)=\{x:f(x)>0\}\cap[-n,n]$ and $A_-(n)=[-n,n]\setminus A_+(n)$. Assuming $f$ is measurable these sets are also measurable and by (3) $$\int_{A_\pm(n)}f(x)dx=0$$ EDIT: and hence $f=0$ almost everywhere.

Please forgive me if I write $dx$ for the Lebesgue measure which I presume is what you refer to as $dm$.

Indeed, as you expected, a simple proof of the result can be found; see Theorem 2.1 in this useful note on absolutely continuous functions.

EDIT: Since this is a quite important result, it is worth giving here the proof in detail. The proof below is essentially the one given in the link above, but somewhat shorter.

Theorem. If $f$ is integrable on $[a, b]$ and $\int_a^x {f(t) dt} = 0$ $\forall x \in [a,b]$, then $f = 0$ a.e. on $[a, b]$.

Proof. An open subset $O$ of $[a,b]$ is a countable union of disjoint open intervals $(c_n, d_n)$; hence, $$ \int_O {f(t) dt} = \sum\limits_{n = 1}^\infty {\int_{c_n }^{d_n } {f(t) dt} } = 0 $$.
If $K$ is a closed subset of $[a,b]$, then $$ \int_K {f(t) dt} = \int_a^b f(t)dt - \int_{(a, b) \setminus K} f(t)dt = 0 - 0 = 0, $$

since $ (a, b) \setminus K $ is open.

Next let $E_ + = \{ x \in [a,b]:f(x) > 0\}$ and $E_ - = \{ x \in [a,b]:f(x) < 0\}$. If $\lambda(E_+) > 0$, then there exists some closed set $K \subset E_+$ such that $\lambda(K) > 0$. But $\int_K {f(t){\rm d}t} = 0$, hence $f=0$ a.e. on $K$. This contradiction shows that $\lambda(E_+) = 0$. Similarly, $\lambda(E_-) = 0$. The theorem is thus established.

I think you can use Dynkin's lemma (if you call this "more elementary").

Let D be all the measurable sets $U\subseteq I=[0,1]$ such that $\intop_U f(t) = 0$ (the function $f\mid_I$ is in $L_2$ so it is also in $L_1$, so I assume this from now). $I\in D$ and if $A\subseteq B\subseteq I$ are in $D$ then $B-A \in D$. If $A_i \subseteq I$ is an increasing sequence in D then $\bigcup A_i \subseteq I$ is also in D (by the DCT). This shows that D is a Dynkin system.

Let P be all the open intervals in I (so $P\subseteq D$). P is not empty and an intersection of two open intervals are open, so P is closed under finite intersection, hence it is a pi system.

Dynkin's lemma says that if P is a pi system and D a dynkin system such that $P\subseteq D$ then $\sigma(P)\subseteq D$. The sigma algebra generated by P is the Borel algebra.

Now look on the set $A=\{x\in I \mid f(x)\geq 0\}$. This is a Lebesgue measurable set, so up to a zero measure set it is Borel measurable set $A'$. Since $\intop_{A'} f(t) = 0$ and f is non negative there, then f is zero almost every where in A'. The same argument work for when f<0, so you get that f is zero almost everywhere in $I$. Now do this for all of $n+I,\;n\in \mathbb{Z}$.