Proof that a standard Brownian motion visits zero infinitely often at the beginning

Let $B_t, t\ge0$ be a standard Brownian motion that starts at 0. Prove that, for any $\epsilon > 0$, $$P(\#\{t\in(0,\epsilon]\mid B_t = 0\}=\infty)=1.$$

This is a well-known fact, but surprisingly I cannot find a single proof on this site. The closest I get is this question, which proves the zero set of a standard Brownian motion restricted to $[0,1]$ is homeomorphic to the Cantor set. However, this proof seems to have utilised unfamiliar properties such as the ternary-expansion construction of the zero set.

Does there exist an elementary proof that only uses the familiar properties of the Brownian motion (normality, independent increments etc.)?

Here's another way. Define $T_a:=\inf\{t>0: B_t=a\}$. It's well known that $E[\exp(-\lambda T_a)]=\exp(-|a|\sqrt{2\lambda})$ for $\lambda>0$ and $a\not=0$. As $a$ decreases to $0$, $T_a$ decreases to $T_{0+}:=\inf\{t>0:B_t>0\}$, and by the Laplace transform cited above, $P[T_{0+}=0]=1$. Similarly, with $T_{0-}:=\inf\{t>0:B_t<0\}$ you have $P[T_{0-}=0]=1$. By path conitnuity and the Intermediate Value Theorem, the zero set $\{t>0: B_t=0\}\cap(0,\epsilon)$ is non-empty for all $\epsilon>0$, with probability 1.

Per @Did's comment, the key is to note that $tB_{1/t}$ also defines a standard BM. And thus $$P(\#\{t\in(0,\epsilon]\mid B_t = 0\}=\infty)=P(\#\{t\geqslant1/\epsilon\mid B_t = 0\}=\infty)$$ and what's left is to show $B_t$ visits $0$ i.o. in $t\ge M$ for any $M>0$.

Here's how I proved it: consider the "first return time" $$\tau:=\inf\{t\color{red}{>} 0\mid B_t=0\}$$ Then it suffices to show $\tau<\infty$ a.s.

Now set any $h>0$. Note that among all the possibilities that $B_t$ returns to 0, one special possibility is that $B_t$ first hits $+h$ and then moves $-h$ to hit $0$. We investigate this special possibility. Define $\tau_1 = \{t>0\mid B_t = h\}$ and $\tau_2$ to be the time it takes $B_t$ to hit 0 starting from $+h$. Clearly $\tau_{1,2}$ are independent and $\tau_2$ equals $\tau_3:=\{t>0\mid B_t=-h\}$ in distribution.

Since "first hitting $+h$ then hitting $0$" is just one among many ways to return to $0$, we have $\tau\le\tau_1+\tau_2$. Therefore $$P(\tau<\infty)\ge P(\tau_1+\tau_2 <\infty)=P(\tau_1<\infty)P(\tau_2<\infty)=P(\tau_1<\infty)P(\tau_3<\infty)=1.$$