This can be shown using the concept of oscillation.
For a bounded $f$, the oscillation of $f$ over set $A$ (which is not a single point) is given by
$$w(A) = \sup_{A} f - \inf_{A} f$$
For a single point $x$, the oscillation is defined as
$$ w(x) = \inf_{J} w(J)$$
where $J$ ranges over bounded intervals containing $x$.
Note that if $x \in I$, then $w(x) \le w(I)$.
Now if $f$ is Riemann integrable over $[a,b]$, then we can show that given any $n \gt 0$, there is a sub-interval $I_{n}$ of $[a,b]$ such that $w(I_n) \le \frac{1}{n}$.
This is because, if every subinterval $I$ of $[a,b]$ had $w(I) \gt \frac{1}{n}$, then for every partition of $[a,b]$ the difference between the upper and lower sums would be at least $\frac{b-a}{n}$ and as a consequence, $f$ would not be integrable.
Now pick $I_{n+1} \subset I_{n}$ such that $w(I_{n+1}) \le \frac{1}{n+1}$.
By completeness there is a point $c$ such that $c \in I_n$.
Thus $w(c) \le w(I_n) \le \frac{1}{n}$ for all $n$. Hence $w(c) = 0$.
Now it can be show that $f$ is continuous at point $x$ if and only if $w(x) = 0$.
Note: This is basically a simplification of one proof of the Riemann Lebesgue theorem of continuity almost everywhere.
Your proof is valid for a bounded function defined on the closed, bounded interval $[a,b]$, despite the apparent simplicity. It also relies on the fact that the Riemann and Lebesgue integrals are the same for step functions, as mentioned by Tony Piccolo.
The other proof you mention is also valid but takes you on a more roundabout path because it strings together a number of results, each of which is not altogether trivial to prove.
A bounded, measurable function defined on a set of finite
measure is Lebesgue integrable.
If a sequence of measurable functions converges almost everywhere
to $f$, then the limit function $f$ is measurable.
If $f$ is Riemann integrable on $[a,b]$, then there exists a
sequence of simple (measurable) functions converging almost everywhere
to $f$.
Adding even more complexity, the proof of the third statement that I know also uses the fact that the set of discontinuities of a Riemann integrable function must be of measure zero. It begins by constructing a partition of $[a,b]$ with dyadic intervals:
$$I_{n,k} = \begin{cases}[a + (b-a)\frac{k-1}{2^n}, a + (b-a)\frac{k}{2^n})\,\,\, k = 1 , \ldots, 2^n -1 \\ [a + (b-a)\frac{2^n-1}{2^n}, b] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, k = 2^n\end{cases} $$
With $m_k = \inf_{I_{n,k}}f(x)$, we can construct the sequence of simple functions
$$\phi_n(x) = \sum_{k=1}^{2^n} m_k \, \chi_{I_{n,k}}(x)$$
The sequence is increasing and, since $f$ is bounded, by monotonicity is convergent to some function $\phi$ such that $\phi_n(x) \uparrow \phi(x) \leqslant f(x).$ If $x \in I_{n,k}$ then the oscillation of $f$ over that interval satisfies
$$\sup_{u,v \in I_{n,k}}|f(u) - f(v)| \geqslant f(x) - \phi_n(x) \geqslant f(x) - \phi(x).$$
Thus, $f(x) \neq \phi(x)$ only at points where $f$ is not continuous, which must belong to a measure zero set if $f$ is Riemann integrable. Therefore, $f$ is almost everywhere the limit of a sequence of simple functions and is measurable.
I would agree that the second approach is "pretty complicated". However, I think it is not uncommon to find theorems with a variety of proofs ranging in complexity.
Best Answer
Let's use the Darboux definition of the Riemann integral.
Let $\epsilon, \delta > 0$. There must exist a partition $P = \{a = x_0 < x_1 < \dots < x_n = b\}$ for which the upper sum $U_{f,P}$ is less than $\epsilon \delta$. Now each interval $[x_{i-1}, x_i]$ on which $f$ attains a value larger than $\delta$ contributes at least $\delta (x_i - x_{i-1})$ to the upper sum. We conclude that the total length of those intervals is at most $\epsilon$. In other words, the set $\{f > \delta\}$ is covered by a finite number of intervals with total length at most $\epsilon$, so its Lebesgue (outer) measure is at most $\epsilon$. But $\epsilon$ was arbitrary, so $\{f > \delta\}$ has Lebesgue measure zero.
Now $\delta$ was also arbitrary, so taking $\delta = 1/k$, we have that $\{f > 0\} = \bigcup_k \{f > 1/k\}$ is a countable union of measure zero sets, hence has measure zero. This needs only the countable subadditivity of Lebesgue measure which is elementary to prove. Or to proceed more directly, fix $\eta > 0$ and use the previous construction with $\epsilon = \eta \cdot 2^{-k}$ to cover the set $\{f > 1/k\}$ by finitely many intervals of total length at most $\eta \cdot 2^{-k}$. Unioning over $k$, we have a covering of $\{f > 0\}$ by countably many intervals of total length at most $\eta$, which by definition of Lebesgue outer measure means that $\{f > 0\}$ has (outer) measure at most $\eta$, and $\eta$ was arbitrary.