= Solution
A left-continuous <adapted process> is a <predictable process>. Define the elementary predictable approximation
$$
H_s^{(n)}=H_{kh_n}\quad\text{for }kh_n<s\le(k+1)h_n,
\qquad h_n=2^{-n}.
$$
At every $s>0$, its sample time approaches $s$ from the left, including when $s$ is itself a grid point. Thus $H_s^{(n)}\to H_s$ pathwise by left continuity.
Use the <continuous semimartingale decomposition> $X=X_0+M+A$, with $M$ a <continuous local martingale> and $A$ a continuous <finite-variation process>. Localize so that on a fixed horizon $T$, the integrands are bounded by a deterministic constant, $[M]_T$ is bounded, and the <total-variation process> of $A$ at $T$ is bounded. This is possible using local boundedness and the finite pathwise clock and variation.
For the finite-variation integral, the <dominated convergence theorem> gives
$$
\sup_{t\le T}\left|\int_0^t(H_s^{(n)}-H_s)\,dA_s\right|
\le\int_0^T|H_s^{(n)}-H_s|\,d|A|_s\longrightarrow0
$$
<almost surely>. For the martingale integral, the <Itô isometry> and the <Doob L2 maximal inequality> give
$$
\mathbb E\sup_{t\le T}\left|\int_0^t(H_s^{(n)}-H_s)\,dM_s\right|^2
\le4\mathbb E\int_0^T|H_s^{(n)}-H_s|^2\,d[M]_s\longrightarrow0.
$$
Here the localized bounds justify dominated convergence also in expectation. Remove localization to conclude that the full elementary integrals converge in <uniform convergence on compacts in probability> to $\int H\,dX$.
The sum of completed intervals differs from the full elementary integral at time $t$ by
$$
H_{h_n\lfloor t/h_n\rfloor}
\left(X_t-X_{h_n\lfloor t/h_n\rfloor}\right).
$$
Its maximum on $[0,T]$ tends to zero <almost surely>, by the local bound on $H$ and <uniform continuity> of $X$. Therefore
$$
\boxed{\sum_{k=0}^{\lfloor t/h_n\rfloor-1}H_{kh_n}
(X_{(k+1)h_n}-X_{kh_n})
\xrightarrow{\mathrm{u.c.p.}}\int_0^tH_s\,dX_s.}
$$
This proves <left-endpoint approximation of a continuous semimartingale integral>, including the unfinished interval convention.
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