Assume $t\ge T$.
$$
[B_t, B_t^T]_{[T,t]} = \int_T^t dB_t dB_t^T = \int_T^t dB_t (dB_t-dB_{t-T}) = \int_T^t (dB_t)^2 - \int_T^tdB_t dB_{t-T}
$$
$$
=\int_T^t dt - \int_T^t dB_t dB_{t-T} = (t-T) - \int_T^t dB_t dB_{t-T} = t-T
$$
using
$$
\int_T^t dB_tdB_{t-T}\approx \sum_{k=0}^{n-1} (B_{t_{k+1}}-B_{t_k})(B_{t_{k+1}-T}-B_{t_k-T})\approx 0$$
since the changes in the product are independent.
It is true that $\langle M\rangle^n_1 \to \langle M\rangle_1$ in $L^1$ when $M$ is a continuous square-integrable martingale. In fact, the result holds even if $M$ is cadlag, as long as $\langle M\rangle$ is continuous.
Indeed, $M^2$ is then a submartingale of class D and so, since
$$E[(M_{t_{i+1}} - M_{t_{i}})^2 |\mathcal{F}_{t_i}]=E[M_{t_{i+1}}^2 - M_{t_{i}}^2 |\mathcal{F}_{t_i}],$$
the result follows from the analogous results for the predictable component of the Doob decomposition of a submartingale of class D, which was proved in
Doléans, Catherine. "Existence du processus croissant naturel associé à un potentiel de la classe (D)." Zeitschrift für Wahrscheinlichkeitstheorie und verwandte Gebiete 9, no. 4 (1968): 309-314.
and which constitutes Theorem 31.2, chapter 6 of
Rogers, L. Chris G., and David Williams. Diffusions, Markov processes and martingales: Volume 2, Itô calculus. Vol. 2. Cambridge university press, 2000.
When $M$ is cadlag (and $\langle M\rangle$ is not continuous), $\langle M\rangle^n_1$ converges to $\langle M\rangle_1$, but only in the $\sigma\left(L^{1}, L^{\infty}\right)$-topology, see for example
Rao, K. Murali. "On decomposition theorems of Meyer." Mathematica Scandinavica 24, no. 1 (1969): 66-78.
In this case, it follows from Mazur's lemma that one can obtain (strong) convergence in $L^1$ by
replacing $(\langle M\rangle^n_1)_n$ with a forward convex combinations thereof. In fact, taking forward convex combinations, one can obtain convergence in $L^1$ simultaneously at all times $t\in [0,1]$, as I showed in
Siorpaes, Pietro. "On a dyadic approximation of predictable processes of finite variation." Electronic Communications in Probability 19 (2014): 1-12.
Best Answer
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 $\hat{T}^{(m)}\wedge T^{(m)}$.