= Solution
At exit, continuity gives $X_T^2=1$. The preceding density identity becomes
$$
Z_T=\exp\left(\frac12-\frac T2-\frac12\int_0^T X_s^2ds\right).
$$
Since $|X_s|\le1$ before exit, $\int_0^TX_s^2ds\le T$. On $\{T\le t\}$ it follows that $Z_T\ge e^{1/2-T}\ge e^{1/2-t}$. The definition of the changed measure now yields
$$
\boxed{\mathbb P(T\le t)
=\widetilde{\mathbb E}[Z_T\mathbf1_{\{T\le t\}}]
\ge e^{1/2-t}\widetilde{\mathbb P}(T\le t).}
$$
The positive sign of the <stochastic integral> in the density is what introduces the positive drift $X_tdt$; reversing that sign would give a different measure and would not prove this inequality.
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