I am trying to find the hole in the proof. What I think it is:
There's a hole in the part where we prove that "$f$ is regulated implies that left and right limits exist". Namely, we forget to prove for the right-sided limit $f(x+)$.
Here is my amendment:
Let $x ∈ (a, b]$ and $x_n → x+$. Let $ϕ_k ∈ S[a, b] $such that $||f − ϕ||_{∞} < 1/k$
For any $k$, there exists $δ_k$ such that $ϕ_k$ is constant on $(x, x+\delta_k)$. Then there exists $N_k$ such that $|x_n −x| < δ_k$ if $n > N_k$, and $$|f(x_m)-f(x_n)| \lt \frac{2}{k}$$
$\forall m,n \gt N_k$. So $(f(x_n))_{n \geq 1}$ is cauchy, hence convergent. Let $f(x+)$ denote its limit.
If $(x^{'}_n)$ is another sequence that converges to $x+$, we can consider the mixed sequence $(x^{"}_n=(x_1,x^{'}_1,x_2,x^{'}_2,...)$. We have $x^{"}_n \rightarrow x+$, so $f(x^{"}_n)$ is Cauchy by the above argument, hence convergent. All its subsequences converge to the same imit, so $f(x^{'}_n)\rightarrow f(x+)$ as well.
Would this be correct?