TheA partial answer is yes.:
By the $L^2$ bound on $\mu$ and Doob's inequality, $M_t$ is $L^2$ bounded, and a fortiori uniformly integrable. Hence it is closable - that is, $M_t = \mathbb E[M_1 | \mathcal M_t]$, where $\mathcal M_t$ is the natural filtration of $M$.
Now let $X$ be such that $\text{Law}(X) = \text{Law}(M)$ and $X_1, Y_1$ be an optimal coupling of $\mu$ and $\nu$. By definition we then have $\mathbb E[|X_1 - Y_1|^2] \leq \varepsilon^2$.
Define the martingale $Y$ by $Y_t := \mathbb E[Y_1| \mathcal X_t]$, where $\mathcal X_t$ is the natural filtration of $X$. Then $X - Y$ is an $\mathcal X_t$ martingale, and we have by Doob's inequality
$$\mathbb E[\sup_{0 \leq t \leq 1} |X_t - Y_t|^2] \leq 4 \mathbb E[|X_1 - Y_1|^2] \leq 4\varepsilon^2,$$
as desired.
To seeIt is left to show that $Y$ is continuous. Unfortunately, weI could only get that for every $T \in [0, 1]$, $Y$ is continuous at $T$ almost surely - but the null sets can depend on $T$, thus it cannot be trivially upgraded to almost sure continuity at all times. I outline the argument for fixed $T$ below.
To get left continuity, apply the almost sure martingale convergence theorem to the $L^2$ boundeduniformly integrable martingale $Y_t$ between $t = 0$ and$Y$ restricted to $T$ for each$[0, T]$. The theorem says that $T$ in$Y_t \to Y_T$ almost surely as $[0, 1]$ to get$t \to T^-$, hence the left continuity.
Meanwhile, and then similarly applyapplying the backward martingale convergence theorem to the $L^2$ bounded process $Z_t := Y_{1-t}$ between $0$ and $1-T$, we get that $Y_t \to Y_T$ almost surely as $t \to T^+$, hence the right continuity.