= Solution
Under the corrected orthogonality hypothesis, the preceding <quadratic variation> calculation gives $[X]_t=A_t=\int_0^te^{2B_s}ds$. We justify the needed <infinite occupation time of one-dimensional Brownian motion>, rather than inferring an infinite integral from recurrent visits alone.
Let $T_0=0$, and recursively let $S_n$ be the first exit of $B$ from $(-1,1)$ after $T_n$, and $T_{n+1}$ its first return to zero after $S_n$. <Recurrence of one-dimensional Brownian motion> makes these <stopping times> finite <almost surely>. The <Strong Markov property> shows that $D_n=S_n-T_n$ are <independent and identically distributed random variables>, each strictly positive. Choose $\varepsilon>0$ with $\mathbb P(D_n\geq\varepsilon)>0$. The <Borel-Cantelli lemma> gives infinitely many such durations and hence $\sum_nD_n=\infty$ <almost surely>. These intervals are disjoint, and $|B_s|\leq1$ on each, so
$$
\boxed{[X]_\infty=\int_0^\infty e^{2B_s}\,ds\geq e^{-2}\sum_{n=0}^\infty D_n=\infty\quad\text{almost surely}.}
$$
This proves the intended conclusion. Without the orthogonality correction, the printed setup gives instead the more general <quadratic variation> in the previous part; it does not justify identifying that quantity with $A$. The proof here is explicitly for the corrected setup.
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