Show that an integratable local martingale $(X_t)_{t=1,...,T}$ is martingale. Is this true in continuous time $[0,\infty)$?
Let $(\tau_n)$ be a localising sequence for X (so $(\tau_n)$ stopping times with $\mathbb{P}(\tau_n=T)\to 1$). I need to show $$E(1_A(X_t-X_s))=0, s\leq t, A \in\mathcal{F}_s.$$ Since $X_{t\land\tau_n}$ is martingale, it holds $$E(1_A(X_{t\land\tau_n}-X_{s\land\tau_n}))=0\ \forall n\in\mathbb{N}, s\leq t, A\in\mathcal{F}_s$$
If $$|1_A(X_{t\land\tau_n}-X_{s\land\tau_n})|\leq\max_{i=1,...,T}|X_i|\leq\sum_{i=1}^T|X_i|<\infty \tag{1}$$ we can apply dominated convergence theorem:
Let $A\in\mathcal{F}_{s\land\tau_n}, s\leq t$, then $$0= lim_{n\to\infty}E(1_A(X_{t\land\tau_n}-X_{s\land\tau_n}))\overset{\text{dom.Konv.}}{=}E(lim_{n\to\infty}1_A(X_{t\land\tau_n}-X_{s\land\tau_n}))=E(1_A(X_{t\land T}-X_{s\land T}))=E(1_A(X_{t}-X_{s})).$$ but is eq. (1) even true? As a hint was given: Show $X_{t\land \tau_n} \to X_t$ by considering $\mathbb{E} \sup_n |X_{t\land \tau_n}|$. I don't know why and how...
Your proof looks good, there are just a few minor details that could be cleaned up:
I think (1) should have a factor of $2$ in it: \begin{align*} |1_A(X_{t \wedge \tau_n} - X_{s \wedge \tau_n})| \le |X_{t \wedge \tau_n}| + |X_{s \wedge \tau_n}| \le 2 \max_{i=1,...,T} |X_i| \le 2 \sum_{i=1}^T |X_i|. \end{align*}
Also, to apply the dominated convergence theorem, we need that $\mathbb{E}[\sum_{i=1}^T |X_i|] < \infty$ rather than just $\sum_{i=1}^T |X_i| < \infty$. Both are true; this is just a minor detail.
This is not true in continuous time. As a counterexample, one can consider the local martingale $M_t := \|B_t\|^{-1}$ where $B_t$ is a 3-dimensional Brownian motion with $B_0 = x \ne 0$. It is shown here that $M$ is a local martingale and is uniformly integrable (in particular, $\mathbb{E}[M_t] < \infty$ for all $t$). However, $M_t \rightarrow 0$ almost surely. If $M$ were a martingale, we would have $M_t = \mathbb{E}[\lim_{s \rightarrow \infty} M_s | \mathcal F_t] = 0,$ which is clearly not true.