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Fix $A>0$; we will be proving that $X_t\stackrel{\mathcal D}= B_t$ on $[0;A].$ Let $T^{(m)}$ be an a.s. increasing sequence of stopping times such that $T^{(m)}\to\infty$ almost surely, and for each $m$, $X_{n\wedge T^{(m)}}$ is a martingale. Then, $Y_n=X_{\epsilon n\wedge T^{(m)}}$, $n=0,1,\dots,$ is a discrete-time martingale satisfying $$\mathbb{E}((Y_{n+1}-Y_n)^2|\mathcal{F}_{\epsilon n})=\mathbb{E}(\epsilon\wedge(T^{(m)}-\epsilon n)_+|\mathcal{F}_{\epsilon n})=:\epsilon_n.$$

Put $t_n:=\sum_{i=0}^{n-1}\epsilon_n$. Note that $0\leq\epsilon_n\leq \epsilon$ almost surely, hence $\epsilon n-t_n$ is non-negative and increasing. We infer that for $N=[A\epsilon^{-1}]$, $$\mathbb{P}(\max_{n\leq N}|\epsilon n-t_n|\geq \delta)=\mathbb{P}(\epsilon N-t_N\geq \delta)\leq \delta^{-1}\mathbb{E}(\epsilon N-t_N)=\delta^{-1}\mathbb{E}((N\epsilon-T^{(m)})_+)\leq\delta^{-1}\mathbb{E}((A+1)-T^{(m)})_+).$$ By monotone convergence, the right-hand side tends to zero as $n\to\infty,$ note that it also does not depend on $\epsilon.$ Therefore, given $\delta>0$, we can choose $m$ such that for all $\epsilon$, $$ \mathbb{P}(\max_{n\leq N}|\epsilon n-t_n|\geq \delta)\leq \delta. $$

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 $\max_{n\leq A\epsilon^{-1}}|\tau_n - n\epsilon|\to 0$ in probability as $\epsilon\to 0$ and then $m\to\infty$. In view of the above discussion, it is enough to show that for any fixed $m$, $\max_{n\leq A\epsilon^{-1}}|\tau_n - t_n|\to 0$ in probability as $\epsilon\to 0$.

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:

$$\mathbb{E}(\tau_{n+1}-\tau_n|\mathcal{F}'_n)=\mathbb{E}(B_{\tau_{n+1}}^2-B_{\tau_n}^2|\mathcal{F}'_n)=\mathbb{E}(Y_{n+1}^2-Y_{n}^2|\mathcal{F}'_n)=\epsilon_n.$$ 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 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 $$ \mathbb{E}(\tau_{N}-t_N)^2\leq \sum_{n=0}^{N-1}\mathbb{E}(\tau_{n+1}-\tau_n)^2\leq \sum_{n=0}^{N-1}\mathbb{E}(Y_{n+1}-Y_{n})^4. $$

It remains to show that the right-hand side tends to zero as $\epsilon\to 0$. This is well-known for bounded, continuous local martingale, see Karatzas-Shreve, Lemma 5.10. We can reduce to this case by localization: since $X_t$ is a.s. bounded on any finite interval, we have that $\hat{T}^{(m)}:=\min\{t:|X_t|\geq m\}\to \infty$ almost surely, we can replace $T^{(m)}$ by $\hat{T}^{(m)}\wedge T^{(m)}$.

Fix $A>0$; we will be proving that $X_t\stackrel{\mathcal D}= B_t$ on $[0;A].$ Let $T^{(m)}$ be an a.s. increasing sequence of stopping times such that $T^{(m)}\to\infty$ almost surely, and for each $m$, $X_{n\wedge T^{(m)}}$ is a martingale. Then, $Y_n=X_{\epsilon n\wedge T^{(m)}}$, $n=0,1,\dots,$ is a discrete-time martingale satisfying $$\mathbb{E}((Y_{n+1}-Y_n)^2|\mathcal{F}_{\epsilon n})=\mathbb{E}(\epsilon\wedge(T^{(m)}-\epsilon n)_+|\mathcal{F}_{\epsilon n})=:\epsilon_n.$$

Put $t_n:=\sum_{i=0}^{n-1}\epsilon_n$. Note that $0\leq\epsilon_n\leq \epsilon$ almost surely, hence $\epsilon n-t_n$ is non-negative and increasing. We infer that for $N=[A\epsilon^{-1}]$, $$\mathbb{P}(\max_{n\leq N}|\epsilon n-t_n|\geq \delta)=\mathbb{P}(\epsilon N-t_N\geq \delta)\leq \delta^{-1}\mathbb{E}(\epsilon N-t_N)=\delta^{-1}\mathbb{E}((N\epsilon-T^{(m)})_+)\leq\delta^{-1}\mathbb{E}((A+1)-T^{(m)})_+).$$ By monotone convergence, the right-hand side tends to zero as $n\to\infty,$ note that it also does not depend on $\epsilon.$ Therefore, given $\delta>0$, we can choose $m$ such that for all $\epsilon$, $$ \mathbb{P}(\max_{n\leq N}|\epsilon n-t_n|\geq \delta)\leq \delta. $$

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 $\{Y_n\}_{n\geq 0}.$ Since both $B_t$ and $X_t$ are continuous, it is enough to show that $\max_{n\leq A\epsilon^{-1}}|\tau_n - n\epsilon|\to 0$ in probability as $\epsilon\to 0$ and then $m\to\infty$. In view of the above discussion, it is enough to show that for any fixed $m$, $\max_{n\leq A\epsilon^{-1}}|\tau_n - t_n|\to 0$ in probability as $\epsilon\to 0$.

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:

$$\mathbb{E}(\tau_{n+1}-\tau_n|\mathcal{F}'_n)=\mathbb{E}(B_{\tau_{n+1}}^2-B_{\tau_n}^2|\mathcal{F}'_n)=\mathbb{E}(Y_{n+1}^2-Y_{n}^2|\mathcal{F}'_n)=\epsilon_n.$$ 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 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 $$ \mathbb{E}(\tau_{N}-t_N)^2\leq \sum_{n=0}^{N-1}\mathbb{E}(\tau_{n+1}-\tau_n)^2\leq \sum_{n=0}^{N-1}\mathbb{E}(Y_{n+1}-Y_{n})^4. $$

It remains to show that the right-hand side tends to zero as $\epsilon\to 0$. This is well-known for bounded, continuous local martingale, see Karatzas-Shreve, Lemma 5.10. We can reduce to this case by localization: since $X_t$ is a.s. bounded on any finite interval, we have that $\hat{T}^{(m)}:=\min\{t:|X_t|\geq m\}\to \infty$ almost surely, we can replace $T^{(m)}$ by $\hat{T}^{(m)}\wedge T^{(m)}$.

Fix $A>0$; we will be proving that $X_t\stackrel{\mathcal D}= B_t$ on $[0;A].$ Let $T^{(m)}$ be an a.s. increasing sequence of stopping times such that $T^{(m)}\to\infty$ almost surely, and for each $m$, $X_{n\wedge T^{(m)}}$ is a martingale. Then, $Y_n=X_{\epsilon n\wedge T^{(m)}}$, $n=0,1,\dots,$ is a discrete-time martingale satisfying $$\mathbb{E}((Y_{n+1}-Y_n)^2|\mathcal{F}_n)=\mathbb{E}(\epsilon\wedge(T^{(m)}-\epsilon n)_+|\mathcal{F}_n)=:\epsilon_n.$$

Put $t_n:=\sum_{i=0}^{n-1}\epsilon_n$. Note that $0\leq\epsilon_n\leq \epsilon$ almost surely, hence $\epsilon n-t_n$ is non-negative and increasing. We infer that for $N=[A\epsilon^{-1}]$, $$\mathbb{P}(\max_{n\leq N}|\epsilon n-t_n|\geq \delta)=\mathbb{P}(\epsilon N-t_N\geq \delta)\leq \delta^{-1}\mathbb{E}(\epsilon N-t_N)=\delta^{-1}\mathbb{E}((N\epsilon-T^{(m)})_+)\leq\delta^{-1}\mathbb{E}((A+1)-T^{(m)})_+).$$ By monotone convergence, the right-hand side tends to zero as $n\to\infty,$ note that it also does not depend on $\epsilon.$ Therefore, given $\delta>0$, we can choose $m$ such that for all $\epsilon$, $$ \mathbb{P}(\max_{n\leq N}|\epsilon n-t_n|\geq \delta)\leq \delta. $$

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 $\max_{n\leq A\epsilon^{-1}}|\tau_n - n\epsilon|\to 0$ in probability as $\epsilon\to 0$ and then $m\to\infty$. In view of the above discussion, it is enough to show that for any fixed $m$, $\max_{n\leq A\epsilon^{-1}}|\tau_n - t_n|\to 0$ in probability as $\epsilon\to 0$.

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:

$$\mathbb{E}(\tau_{n+1}-\tau_n|\mathcal{F}'_n)=\mathbb{E}(B_{\tau_{n+1}}^2-B_{\tau_n}^2|\mathcal{F}'_n)=\mathbb{E}(Y_{n+1}^2-Y_{n}^2|\mathcal{F}'_n)=\epsilon_n.$$ 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 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 $$ \mathbb{E}(\tau_{N}-t_N)^2\leq \sum_{n=0}^{N-1}\mathbb{E}(\tau_{n+1}-\tau_n)^2\leq \sum_{n=0}^{N-1}\mathbb{E}(Y_{n+1}-Y_{n})^4. $$

It remains to show that the right-hand side tends to zero as $\epsilon\to 0$. This is well-known for bounded, continuous local martingale, see Karatzas-Shreve, Lemma 5.10. We can reduce to this case by localization: since $X_t$ is a.s. bounded on any finite interval, we have that $\hat{T}^{(m)}:=\min\{t:|X_t|\geq m\}\to \infty$ almost surely, we can replace $T^{(m)}$ by $\hat{T}^{(m)}\wedge T^{(m)}$.

Fix $A>0$; we will be proving that $X_t\stackrel{\mathcal D}= B_t$ on $[0;A].$ Let $T^{(m)}$ be an a.s. increasing sequence of stopping times such that $T^{(m)}\to\infty$ almost surely, and for each $m$, $X_{n\wedge T^{(m)}}$ is a martingale. Then, $Y_n=X_{\epsilon n\wedge T^{(m)}}$, $n=0,1,\dots,$ is a discrete-time martingale satisfying $$\mathbb{E}((Y_{n+1}-Y_n)^2|\mathcal{F}_{\epsilon n})=\mathbb{E}(\epsilon\wedge(T^{(m)}-\epsilon n)_+|\mathcal{F}_{\epsilon n})=:\epsilon_n.$$

Put $t_n:=\sum_{i=0}^{n-1}\epsilon_n$. Note that $0\leq\epsilon_n\leq \epsilon$ almost surely, hence $\epsilon n-t_n$ is non-negative and increasing. We infer that for $N=[A\epsilon^{-1}]$, $$\mathbb{P}(\max_{n\leq N}|\epsilon n-t_n|\geq \delta)=\mathbb{P}(\epsilon N-t_N\geq \delta)\leq \delta^{-1}\mathbb{E}(\epsilon N-t_N)=\delta^{-1}\mathbb{E}((N\epsilon-T^{(m)})_+)\leq\delta^{-1}\mathbb{E}((A+1)-T^{(m)})_+).$$ By monotone convergence, the right-hand side tends to zero as $n\to\infty,$ note that it also does not depend on $\epsilon.$ Therefore, given $\delta>0$, we can choose $m$ such that for all $\epsilon$, $$ \mathbb{P}(\max_{n\leq N}|\epsilon n-t_n|\geq \delta)\leq \delta. $$

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 $\max_{n\leq A\epsilon^{-1}}|\tau_n - n\epsilon|\to 0$ in probability as $\epsilon\to 0$ and then $m\to\infty$. In view of the above discussion, it is enough to show that for any fixed $m$, $\max_{n\leq A\epsilon^{-1}}|\tau_n - t_n|\to 0$ in probability as $\epsilon\to 0$.

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:

$$\mathbb{E}(\tau_{n+1}-\tau_n|\mathcal{F}'_n)=\mathbb{E}(B_{\tau_{n+1}}^2-B_{\tau_n}^2|\mathcal{F}'_n)=\mathbb{E}(Y_{n+1}^2-Y_{n}^2|\mathcal{F}'_n)=\epsilon_n.$$ 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 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 $$ \mathbb{E}(\tau_{N}-t_N)^2\leq \sum_{n=0}^{N-1}\mathbb{E}(\tau_{n+1}-\tau_n)^2\leq \sum_{n=0}^{N-1}\mathbb{E}(Y_{n+1}-Y_{n})^4. $$

It remains to show that the right-hand side tends to zero as $\epsilon\to 0$. This is well-known for bounded, continuous local martingale, see Karatzas-Shreve, Lemma 5.10. We can reduce to this case by localization: since $X_t$ is a.s. bounded on any finite interval, we have that $\hat{T}^{(m)}:=\min\{t:|X_t|\geq m\}\to \infty$ almost surely, we can replace $T^{(m)}$ by $\hat{T}^{(m)}\wedge T^{(m)}$.

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.

Fix $A>0$; we will be proving that $X_t\stackrel{\mathcal D}= B_t$ on $[0;A].$ Let $T^{(m)}$ be an a.s. increasing sequence of stopping times such that $T^{(m)}\to\infty$ almost surely, and for each $m$, $X_{n\wedge T^{(m)}}$ is a martingale. Then, $Y_n=X_{\epsilon n\wedge T^{(m)}}$, $n=0,1,\dots,$ is a discrete-time martingale satisfying $$\mathbb{E}((Y_{n+1}-Y_n)^2|\mathcal{F}_n)=\mathbb{E}(\epsilon\wedge(T^{(m)}-\epsilon n)_+|\mathcal{F}_n)=:\epsilon_n.$$

Put $t_n:=\sum_{i=0}^{n-1}\epsilon_n$. Note that $0\leq\epsilon_n\leq \epsilon$ almost surely, hence $\epsilon n-t_n$ is non-negative and increasing. We infer that for $N=[A\epsilon^{-1}]$, $$\mathbb{P}(\max_{n\leq N}|\epsilon n-t_n|\geq \delta)=\mathbb{P}(\epsilon N-t_N\geq \delta)\leq \delta^{-1}\mathbb{E}(\epsilon N-t_N)=\delta^{-1}\mathbb{E}((N\epsilon-T^{(m)})_+)\leq\delta^{-1}\mathbb{E}((A+1)-T^{(m)})_+).$$ By monotone convergence, the right-hand side tends to zero as $n\to\infty,$ note that it also does not depend on $\epsilon.$ Therefore, given $\delta>0$, we can choose $m$ such that for all $\epsilon$, $$ \mathbb{P}(\max_{n\leq N}|\epsilon n-t_n|\geq \delta)\leq \delta. $$

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 $\max_{n\leq A\epsilon^{-1}}|\tau_n - n\epsilon|\to 0$ in probability as $\epsilon\to 0$ and then $m\to\infty$. In view of the above discussion, it is enough to show that for any fixed $m$, $\max_{n\leq A\epsilon^{-1}}|\tau_n - t_n|\to 0$ in probability as $\epsilon\to 0$.

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:

$$\mathbb{E}(\tau_{n+1}-\tau_n|\mathcal{F}'_n)=\mathbb{E}(B_{\tau_{n+1}}^2-B_{\tau_n}^2|\mathcal{F}'_n)=\mathbb{E}(Y_{n+1}^2-Y_{n}^2|\mathcal{F}'_n)=\epsilon_n.$$ 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 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 $$ \mathbb{E}(\tau_{N}-t_N)^2\leq \sum_{n=0}^{N-1}\mathbb{E}(\tau_{n+1}-\tau_n)^2\leq \sum_{n=0}^{N-1}\mathbb{E}(Y_{n+1}-Y_{n})^4. $$

It remains to show that the right-hand side tends to zero as $\epsilon\to 0$. This is well-known for bounded, continuous local martingale, see Karatzas-Shreve, Lemma 5.10. We can reduce to this case by localization: since $X_t$ is a.s. bounded on any finite interval, we have that $\hat{T}^{(m)}:=\min\{t:|X_t|\geq m\}\to \infty$ almost surely, we can replace $T^{(m)}$ by $\hat{T}^{(m)}\wedge T^{(m)}$.