I've found a proof that can be reduced to the (incorrect) statement that
$$ P\left( \inf_{t > 0} B_t < 0 \right) = 0 $$
for any Brownian motion $B_t$ started from some positive value, i.e., $B_0 > 0$. This is incorrect, since Brownian motion has finite hitting times for some negative value is finite.
I've tried going through the proof, but each step seems to make sense to me and I'm not sure where the mistake lies. My suspicion is that the final step is faulty, but I would appreciate some help.
Here is how the (faulty) proof of the above incorrect statement, which lower-bounds $B_t$ by a supermartingale, then uses Doob's Martingale Inequality to bound the infimum of the process, then finally takes limits.
Doob's (Super-)Martingale Inequality:
Let $X_t$ be a supermartingale with almost-surely cadlag sample paths and $C > 0$. Then, $$ P(\inf_{0 \leq t \leq T} X_t \leq -C) \leq \frac{E[ \max(-X_T, 0) ]}{C} $$
We now proceed to the main proof.
Proof: Let $T > 0, \epsilon > 0, \delta \in (0, 1)$ be arbitrary. We will first show that $$ P( \inf_{0 \leq t \leq T} B_t < -\epsilon) < \delta. $$ where $B_t$ is a Brownian motion started from $B_0 > 0$. First, define $\theta = \min( \frac{\delta \epsilon}{2T}, B_0)$, so that $\theta \leq B_0$.
Define the sequence of (stopping times) downcrossings $\eta_i$ and upcrossings $\zeta_i$ for $i=0, 1, \dots$ as
$$ \begin{aligned} \eta_0 &= 0, \\ \zeta_0 &= \inf \{ t : B_t > \theta \}, \\ \eta_i &= \inf \{ t : B_t < \theta, t > \xi_{i-1} \}, i=1, 2, \dots, \\ \zeta_i &= \inf \{ t : B_t > \theta, t > \eta_{i-1} \}, i=1, 2, \dots. \end{aligned} $$ Now, define the random process $U_t$ as $$ U_t = \theta + \sum_{i=0}^\infty \left( \int_{\eta_i \land t}^{\zeta_i \land t} -\theta \, d\tau + \int_{\eta_i \land t}^{\zeta_i \land t} dB_\tau \right) $$ In other words, when $U_t$ has a drift of $-\theta$ when $B_t < \theta$. $U_t$ is a supermartingale, since for $s \leq t$, $$ E[U_t | U_s] = U_s + E\left[ \sum_{i=0}^\infty \int_{\eta_i \land t}^{\zeta_i \land t} -\theta \, d\tau \right] \leq U_s. $$
Now, we show by induction that $B_t \leq U_t$ and $U_t \leq \theta$ for all $t \geq 0$. The base case holds, since $U_0 = \theta \leq B_0$ by construction.
Case 1: Suppose the result holds up to time $t \in [\eta_i, \zeta_i], i \geq 0$. Then, since $$ 0 \leq -\theta $$ the (zero) drift of $B_t$ is an upper-bound for the negative drift of $U_t$, while the two terms have the same diffusion terms. Hence, $B_t \geq U_t$ for the entire $[\eta_i, \zeta_i]$. Moreover, since $B_t \leq \theta$ on this interval by definition of $\eta_i, \zeta_i$, we have $U_t \leq \theta$ as well.
Case 2: Suppose the result holds up to time $t \in [\zeta_i, \eta_{i+1}], i \geq 0$. Then, by definition of $\zeta_i$, $$ U_t = U_{\zeta_i} \leq B_{\zeta_i} = \theta \leq B_t. $$ Combining the two cases then gives us that $B_t \leq U_t$ and $U_t \leq \theta$ for all $t \geq 0$. Since $U_t \leq B_t$ for $t \geq 0$, this implies that $$ P\left( \inf_{0 \leq t \leq T} B_t < -\epsilon\right) \leq P\left( \inf_{0 \leq t \leq T} U_t < -\epsilon \right) $$ Since $U_t$ is a supermartingale, applying Doob's Martingale Inequality then gives us $$ P\left( \inf_{0 \leq t \leq T} U_t < -\epsilon \right) \leq \frac{E[\max(-U_T, 0)]}{\epsilon}. $$ We know that $$ E[U_T] = \theta + E\left[ \sum_{i=0}^\infty \int_{\eta_i \land t}^{\zeta_i \land t} -\theta \, d\tau \right] \geq \theta - \theta T $$ We also know that $$U_T \leq \theta$$ almost surely, hence $$E[\max(U_T, 0)] \leq \theta$$ Since $$E[\max(-U_T,0)] = E[\max(U_T,0)] - E[U_T],$$ combining the above two statements gives us that $$ E[\max(-U_T, 0)] \leq \theta -\theta + \theta T = \theta T $$ which gives us $$ \begin{aligned} P\left( \inf_{0 \leq t \leq T} B_t < -\epsilon \right) &\leq \frac{\theta T}{\epsilon} = \frac{\delta \epsilon}{2T} \frac{T}{\epsilon} < \delta \end{aligned} $$ Since $T, \delta, \epsilon$ were arbitrary, taking the limit of $T \to \infty, \delta \to 0$ and $\epsilon \to 0$ then "proves" that $$ P\left( \inf_{t \geq 0} B_t < 0 \right) = 0 \tag*{$\blacksquare$} $$