Let $(\sigma_n)_{n \in \mathbb{N}}$ be a localizing sequence of the local martingale $M$, i.e. an increasing sequence of stopping times such that $\sigma_n \to \infty$ and $(M_{t \wedge \sigma_n})_{t \geq 0}$ is a martingale for each $n \in \mathbb{N}$.
Applying the optional stopping theorem, we find that $(M_{t \wedge \sigma_n \wedge T(s)})_{t \geq 0}$ is a martingale, i.e.
$$\mathbb{E}(M_{t \wedge \sigma_n \wedge T(s)} \mid \mathcal{F}_u) = M_{u \wedge \sigma_n \wedge T(s)} \tag{1}$$
for all $u \leq t$. Since $M$ has continuous sample paths, the right-hand side converges to $M_{u \wedge T(s)}$ as $n \to \infty$. If we can show that
$$\lim_{n \to \infty} \mathbb{E}(M_{t \wedge \sigma_n \wedge T(s)} \mid \mathcal{F}_u) = \mathbb{E}(M_{t \wedge T(s)} \mid \mathcal{F}_u)\tag{2}$$
this proves that $(M_{t \wedge T(s)})_{t \geq 0}$ is a martingale. To prove $(2)$, we note that by Doob's maximal inequality
$$\begin{align*} \mathbb{E} \left( \sup_{r \leq t} |M_{r \wedge \sigma_n \wedge T(s)}|^2 \right) &\leq 4 \mathbb{E}(M_{t \wedge \sigma_n \wedge T(s)}^2) = 4 \mathbb{E} \big( [M]_{t \wedge \sigma_n \wedge T(s)} \big) \leq \mathbb{E}([M]_{T(s)}) \leq 4s. \end{align*}$$
By the monotone convergence theorem this implies
$$\mathbb{E} \left( \sup_{r \leq t} |M_{r \wedge T(s)}|^2 \right) \leq 4s < \infty.$$
Hence,
$$\mathbb{E} \left( \sup_{n \in \mathbb{N}} |M_{t \wedge \sigma_n \wedge T(s)}| \right) \leq \sqrt{\mathbb{E} \left( \sup_{r \leq t} |M_{r \wedge T(s)}|^2 \right)} < \infty.$$
Now $(2)$ follows from the dominated convergence theorem.
It is the collection $\{X^{T_n}_t: t\ge 0\}$ that is uniformly integrable, for each fixed $n$. This doesn't give $L^1$ convergence of $X^{T_n}_t$ to $X_t$.
Your expectation computation would work if you knew that the collection $\{X^{T_n}_t: n=1,2,\ldots\}$ was uniformly integrable, for fixed $t$.
Best Answer
For $n \in \mathbb{N}$ define a stopping time $\tau_n$ by $$\tau_n := \inf\{t>0; |X_t| \geq n\}.$$ Since $X$ has continuous sample paths, it holds that $|X_{t \wedge \tau_n}| \leq n$ for all $t \geq 0$. The sequence
$$t_n' := \min\{t_n, \tau_n\}$$
is a sequence of non-decreasing stopping times such that $(X_{t \wedge t_n'})_{t \geq 0}$ is a continuous martingale (by the optional stopping theorem) and $$\mathbb{E}|X_{t \wedge t_n'}|^2 \leq n^2< \infty.$$