How can I prove this random process to be Standard Brownian Motion?

For every nonnegative $t$, let $Z_t=B_{1-\mathrm e^{-t}}=\displaystyle\int_0^{1-\mathrm e^{-t}}\mathrm dB_s$. Then $(Z_t)_{t\geqslant0}$ is a Brownian martingale and $\mathrm d\langle Z\rangle_t=\mathrm e^{-t}\mathrm dt$ hence there exists a Brownian motion $(\beta_t)_{t\geqslant0}$ starting from $\beta_0=0$ such that $Z_t=\displaystyle\int_0^t\mathrm e^{-s/2}\mathrm d\beta_s$ for every nonnegative $t$. In particular, $X_t=\displaystyle\mathrm e^{t/2}\int_0^t\mathrm e^{-s/2}\mathrm d\beta_s$ and $$ \int_0^tX_u\mathrm du=\int_0^t\mathrm e^{u/2}\int\limits_0^u\mathrm e^{-s/2}\mathrm d\beta_s\mathrm du=\int_0^t\mathrm e^{-s/2}\int_s^t\mathrm e^{u/2}\mathrm du\mathrm d\beta_s, $$ hence $$ \int_0^tX_u\mathrm du=\int_0^t\mathrm e^{-s/2}2(\mathrm e^{t/2}-\mathrm e^{s/2})\mathrm d\beta_s=2\mathrm e^{t/2}\int_0^t\mathrm e^{-s/2}\mathrm d\beta_s-2\beta_t=2X_t-2\beta_t. $$ This proves that $Y_t=X_t-\displaystyle\frac12\int\limits_0^tX_u\mathrm du=\beta_t$ and that $(Y_t)_{t\geqslant0}$ is a standard Brownian motion.