Q) If $T_0 = \inf\{t>0:B_t=0\}$, then prove that $P_0(T_0 = 0)=1$.
There is a similar theorem which states that if $\tau = \inf\{t\geq 0; B_t>0\}$, then $P_0(\tau=0)=1$ whose proof is:
$$P_0(\tau\leq t)\geq P_0(B_t>0)=1/2 \text{ and } P_0(\tau = 0)=\lim_{t\downarrow 0}P_0(\tau\leq t)\geq 1/2$$
and since $\{\tau=0\}\in\mathcal{F}_0^+$- a trivial sigma-field, thus $P_0(\tau=0)=1$.
But I'm not sure how to use a similar argument for the question in hand. Appreciate a hint.
The result you already proved shows that the following holds with probability $1$: there is a sequence $t_n$ of positive numbers decreasing to $0$ such that $B_{t_n} >0$ for all $n$. By symmetry of Brownian motion we see that with probability $1$ there is a sequence $s_n$ of positive numbers decreasing to $0$ such that $B_{s_n} <0$ for all $n$. By continuity of Brownian paths it follows that with probability $1$ there is a sequence $u_n$ of positive numbers decreasing to $0$ such that $B_{u_n} =0$ for all $n$. This finishes the proof.