Can finite variation martingales be approximated by martingales of integrable variation? 
Note: This was a genuine question when I started writing it. Shortly after that, I figured out the answer. I decided to post the answer along with the question because someone else may find this useful. Any feedback on the answer or alternative approaches would be appreciated.

Cohen and Elliott (2015) establish a theorem for finite-variation square-integrable martingales by first proving the result for (square-integrable) martingales of integrable variation. They then appeal to continuity to extend the result to martingales of finite variation.
This argument relies on being able to approximate such finite-variation process with ones of integrable variation. How does one do this? By approximation, I mean this in the $\mathcal H^2$-norm, as defined below.

Definitions and Context
Fix a filtered probability space $(\Omega,\mathcal F,\{\mathcal F_t\}_{t\ge0},P)$ satisfying the usual conditions. That is, the filtration is complete and right-continuous. A martingale $M$ is square-integrable whenever $$ \| M \|_{\mathcal H^2} = \left(E\left[\sup_t |M_t|^2 \right]\right)^{\frac 12} < \infty. $$ We call the space of square-integrable martingales $\mathcal H^2$. We know that $\mathcal H^2$ is a Banach space, subject to the usual identification of indistinguishable martingales.
Say that $M$ has finite variation if it is càdlàg, and if, for each $t\in[0,\infty)$, $$ D^M_t:= \int_{[0,t]} \left|\mathrm d M_s \right|= \sup \left( \sum_i \left| M_{t_{i+1}} -M_{t_i} \right|\right) < \infty \quad \text{a.s.} $$ where the supremum is taken over the increasing deterministic sequences $\left\{t_i\right\}_{i\in\mathbb N}$ in $[0,t]$.
A càdlàg martingale $M$ has integrable variation whenever $$ E\left[D_\infty^M\right] = E \left[ \int_{[0,\infty)}\left|\mathrm d M_s \right| \right] < \infty. $$
Clearly, any process of integrable variation has finite variation, but the converse is not true.
The proof I reference above asserts that for each $M \in \mathcal H^2$ with finite variation, there must exist a sequence $\{M^n\}_{n\in\mathbb N}\subset \mathcal H^2$ where each $M^n$ has integrable variation such that $$ \left\| M^n - M \right\|_{\mathcal H^2} \to 0. $$ Equivalently, one can also show that $$ \left\| M^n_\infty - M_\infty \right\|_{L^2} \to 0, $$ where $\| \cdot \|_{L^2}$ is the $L^2$-norm, as this would imply that $M^n \to M$ in $\mathcal H^2$.
I don't see how to construct such a sequence, or even to show that one exists. I know that the bounded martingales are dense in $\mathcal H^2$, so we can approximate any martingale with a sequence of bounded martingales. However, I don't know how to guarantee that the approximating sequence has integrable variation even though the martingale we wish to approximate has bounded variation. 
Any suggestions?
 A: Yes, but in $\mathcal H^1$, and not necessarily in $\mathcal H^2$. This is sufficient for the theorem referenced in the question.
To prove this, we will need two results. Before that, we start with a definition.
We say that a martingale $X$ is in $\mathcal H^1$ whenever $$ \| X \|_{\mathcal H^1} = E\left[ \sup_t \left|X_t\right|\right] <\infty. $$ Note that martingales in $\mathcal H^1$ are uniformly integrable.

Let $X\in\mathcal H^1$, and suppose that $\{T_n\}_{n\in\mathbb N}$ is an increasing sequence of stopping times with $T_n \to \infty$ a.s. Then, the stopped processes $X^{T_n}$ converge to $X$ in $\mathcal H^1$ as $n\to\infty$.

Proof: By Jensen's inequality, $\left| X_t - X_t^{T_n} \right|$ is a submartingale. Moreover, $$\sup_t \left| X_t - X_t^{T_n} \right| \le 2 \sup_t \left| X_t \right|$$ is integrable since $X\in\mathcal H^1$. Doob's Martingale Inequality gives us that $$ \lambda P\left( \sup_t \left| X_t - X_t^{T_n} \right| > \lambda \right) \le E\left[\left| X_\infty -X_\infty^{T_n} \right|\right]. $$ Since $X_{T_n}\to X_\infty$ a.s., the right-hand side goes to zero by dominated convergence. Thus, $$ \sup_t \left| X_t - X_t^{T_n} \right| \to 0 \qquad \text{in probability}, $$ and the claim follows by dominated convergence. $\blacksquare$
The next result states that martingales of finite variation are of locally integrable variation.

Let $X$ be a square-integrable martingale of finite variation. Then, there exists a nondecreasing sequence of stopping times $\{T_n\}_{n\in\mathbb N}$ with $T_n \to \infty$ a.s. such that $X^{T_n}$ has integrable variation.

Proof: Let $$ T_n = \inf \left\{ t : \int_{[0,t]} \left| \mathrm d X_s \right| \ge n \right\}. $$ $T_n$ is a stopping time with $T_n \uparrow \infty$. By construction we know that $$ \int_{\left[0,T_n\right)} \left| \mathrm d X_s \right| \le n \quad \text{a.s.} $$ Since $X$ is in $\mathcal H^2$, $\left|\Delta X_{T_n}\right| = \left|X_{T_n}-X_{T_n-}\right|$, where $X_{t-} = \lim_{s\uparrow t} X_s$, is integrable. Thus, $$ \int_{\left[0,T_n\right]} \left| \mathrm d X_s \right| \le n + \left| \Delta X_{T_n} \right| $$ is also integrable. $\blacksquare$
