Let $T^{(m)}$ be a stopping time such that $X_{n\wedge T^{(m)}}$ is a martingale (eventually, $T^{(m)}\to\infty$ a.s.). Then, $Y_n=X_{\epsilon n\wedge T^{(m)}}$, $n=0,1,\dots,$ is a discrete-time martingale satisfying $$\mathbb{E}(Y_{n+1}-Y_n)^2=\mathbb{E}(\epsilon\wedge(T^{(m)}-\epsilon n)_+)=:\epsilon_n$$ 

By Skorokhod embedding theorem *[A. V. Skorohod, Studies in the theory of random processes, Addison-Wesley, Reading, Mass.,1965]*, there exists a Brownian motion $\{B_t\}_{t\geq 0}$, which we can take independent of $\{X_t\}_{t\geq 0}$, and a sequence of stopping times $0=\tau_0<\tau_1<\tau_2<\dots$ such that $\{B_{\tau_n}\}_{n\geq 0}$ has the same distribution as $\{Y_n\}_{n\geq 0},$ and $\mathbb{E}(\tau_{n+1}-\tau_n)=\epsilon_n$. Since both $B_t$ and $X_t$ are continuous, it is enough to show that for each $A>0$, $\max_{n\leq A\epsilon^{-1}}|\tau_n - n\epsilon|\to 0$ in probability as $\epsilon\to 0$. Since eventually $T^{(m)}\to \infty$ a.s., this is the same as showing that $\max_{n\leq A\epsilon^{-1}}|\tau_n - t_n|$ is small with high probability, where $t_n=\mathbb{E}\tau_n=\sum_{i=0}^{n-1}\epsilon_n$.

Recall that the construction in Skorokhod embedding theorem runs iteratively: conditionally on $\mathcal{F'_n}=\sigma(\mathcal{F}_{\epsilon n},\tau_n, \{B_{t}\}_{0\leq t\leq\tau_n})$, we construct $\tau_{n+1}-\tau_{n}$ as a stopping time for the Brownian motion $B_{\tau_n+t}-B_{\tau_n}$ such that $B_{\tau_{n+1}}-B_{\tau_n}$ has the same distribution as the conditional distribution of $Y_{n+1}-Y_{n}$. This implies:

$$\mathbb{E}(\tau_{n+1}-\tau_n-\epsilon_n|\mathcal{F}'_n)=\mathbb{E}(B_{\tau_{n+1}}^2-B_{\tau_n}^2-\epsilon_n|\mathcal{F}'_n)=\mathbb{E}(Y_{n+1}^2-Y_{n}^2-\epsilon_n|\mathcal{F}'_n)=0.$$ That is, $\tau_n-t_n$ is in fact a martingale. In particular, by Doob's inequality, we have $$\mathbb{P}(\max_{n\leq N}|\tau_n-t_n|>\delta)\leq\delta^{-2}\mathbb{E}(\tau_N-t_N)^2,$$ which we will apply, say, to $N=[A\epsilon^{-1}]$. In fact, Skorokhod embedding theorem guarantees that 
$\mathbb{E}(\tau_{n+1}-\tau_n)^2\leq C\mathbb{E}(Y_{n+1}-Y_{n})^4$, thus
$$
\mathbb{E}(\tau_{N}-t_N)^2\leq \sum_{n=0}^{N-1}\mathbb{E}(\tau_{n+1}-\tau_n)^2\leq \sum_{n=0}^{N-1}\mathbb{E}(Y_{n+1}-Y_{n})^4.
$$

So, we need to show that the right-hand side tends to zero. In fact, [Karatzas-Shreve, Lemma 5.10] shows that this is the case for any bounded, continuous local martingale. We can reduce to this case by localization: we first choose a sequence of stopping times $T^{(m)}\to \infty$ a.s. such that $X_{t\wedge T^{(m)}}$ are bounded. Then, given $\delta>0,$ we choose $m$ so large that $\sum_{n\leq A\epsilon^{-1}} (\epsilon-\epsilon_n)\leq \mathbb{P}(T^{(m)}<A)<\delta$, then, using the lemma in Karatzas-Shevre, we choose $\epsilon$ so small that $\mathbb{E}(\tau_{N}-t_N)^2<\delta^3$. With this choice, we get 
$$
\mathbb{P}(\max_{n\leq N}|\tau_n-n\epsilon|>2\delta)\leq 2\delta,
$$
which concludes the proof.