= Solution
Take $L>0$, as needed for a nontrivial interval. Since $X$ is continuous and <adapted>, for every level $a$ and $t\geq0$,
$$
\{T_a\leq t\}
=\left\{\inf_{0\leq s\leq t}|X_s-a|=0\right\}.
$$
The infimum is attained on this compact interval and can be computed on the countable dense set $(\mathbb Q\cap[0,t])\cup\{t\}$. It is therefore $\mathcal F_t$-measurable. Thus $T_a$ is a <stopping time>, and so is $T=T_{-L}\wedge T_L$, because $\{T\leq t\}=\{T_{-L}\leq t\}\cup\{T_L\leq t\}$.
By continuity, $X_{t\wedge T}\in[-L,L]$, even before knowing that $T$ is finite. Hence
$$
e^{-L}\leq M_{t\wedge T}\leq e^L.
$$
Stopping preserves the <local martingale> property. The <bounded local martingale criterion> now shows that $M^T=(M_{t\wedge T})_{t\geq0}$ is a <uniformly integrable martingale>; its deterministic bound also gives square integrability. In particular,
$$
\boxed{\mathbb EM_{t\wedge T}=1,\qquad
\mathbb EM_{t\wedge T}^2\leq e^{2L}.}
$$
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