Let $f: [0,1] \to \mathbb{R}$ be a continuous function. Since $[0,1]$ is compact, $f$ is uniformly continuous on $[0,1]$, i.e. we can choose $n \in \mathbb{N}$ such that
$$|f(s)-f(t)| < \frac{\varepsilon}{2} \quad \text{for all $|s-t| \leq \frac{1}{n}$.}$$
If we set $t_j := j/n$ for $j=0,\ldots,n$, then
$$\begin{align*} \mathbb{P} \left( \sup_{0 \leq t \leq 1} |B_t-f(t)| <\varepsilon\right) &\geq \mathbb{P}\left( \forall j=0,\ldots,n-1: \sup_{t \in [t_j,t_{j+1}]} |B_t-f(t_j)| < \frac{\varepsilon}{2 (n-j)} \right) \\ &= \mathbb{P}\left( \prod_{j=0}^{n-1} 1_{A_j} \right) \end{align*}$$
for
$$A_j := \left\{\sup_{t \in [t_j,t_{j+1}]} |B_t-f(t_j)| < \frac{\varepsilon}{2(n-j)} \right\} \in \mathcal{F}_{t_{j+1}}.$$
It follows from the Markov property (of Brownian motion) and tower property (of conditional expectation) that
$$\begin{align*} \mathbb{P}\left( \prod_{j=0}^{n-1} 1_{A_j} \right) &= \mathbb{P} \left[ \left(\prod_{j=0}^{n-2} 1_{A_j} \right) \mathbb{P}^{B_{t_{n-1}}} \left(\sup_{t \in [0,1/n]} |B_t-f(t_{n-1})| < \frac{\varepsilon}{2} \right) \right]. \end{align*}$$
It suffices to show that
$$\mathbb{P}^x \left( \sup_{t \in [0,1/n]} |B_t-f(t_{n-1})| < \frac{\varepsilon}{2} \right)>c>0 \tag{1}$$
for all $x \in B(f(t_{n-1}),\varepsilon/4)$. (Then we can iterate the procedure and obtain the desired lower bound.) To this end, we note that
$$\begin{align*} \mathbb{P}^x \left( \sup_{t \leq 1/n} |B_t-f(t_{n-1})| < \frac{\varepsilon}{2} \right) &= \mathbb{P} \left( \sup_{t \leq 1/n} |B_t+x-f(t_{n-1})| < \frac{\varepsilon}{2} \right) \\ &\geq \mathbb{P} \left( \sup_{t \leq 1/n} |B_t| < \frac{\varepsilon}{4} \right) \end{align*}$$
for all $x \in B(f(t_{n-1}),\varepsilon/4)$. As $M_{1/n} := \sup_{t \leq 1/n} B_t \sim |B_{1/n}|$ (by the reflection principle), $(1)$ follows.
Remark: The asymptotics of the probability $\mathbb{P} \left( \sup_{t \in [0,1]} |B_t-f(t)| < \varepsilon \right)$ as $\varepsilon \to 0$ is subject of so-called small ball estimates.
$$\mathbb{P}(B^x_t\in A)=\mathbb{P}(B^x_t\in A,B^x_s>0,\forall s\in [0,t])+\mathbb{P}(B^x_\tau=0,B^x_t\in A)$$
where $$\tau=inf\{s,B^x_s=0,s\in[0,t]\}$$
And
$$\mathbb{P}(B^x_\tau=0,B^x_t\in A)=\mathbb{P}(B^x_\tau=0,B^x_t\in -A)=\mathbb{P}(B^x_t\in -A)$$
Best Answer
This question is not really about Brownian motion.
You see, $t^*$ and $y$ are independent, which makes: $$\mathbb{E}( y(t^*)^2 |t^*=t )= \mathbb{E}( y(t)^2 |t^*=t)= t$$ Hence forth what follows.
P/s: $t^*$ and $y$ are independent