In a paper I am reading the authors claim that, if $B$ is a standard BM in $\mathbb{R}$ and $f\in C([0,1],\mathbb{R})$, then for any $\epsilon>0$ $$ \mathbb{P}(\sup_{t\in [0,1]}|B_t-f(t)|<\epsilon)>0. $$ They do not give a proof for this result so I assume it is somewhat well-known, but I don't seem to be able to prove it. Looking online, I found that this is Exercise $1.8$ in the book of Morters and Peres (https://people.bath.ac.uk/maspm/book.pdf). They give the following hint: Given $f\in C[0,1]$ and $\epsilon>0$ there exists n such that the function $g\in C[0,1]$, which agrees with $f$ on the dyadic points in $D_n$ and is linearly interpolated inbetween, satisfies $\sup|f(t)−g(t)|<\epsilon$. Then use Lévy’s construction of Brownian motion and the fact that normal distributions have full support to complete the proof.
The hint however seems not to work. In facts, when I approximate $B$ with an appropriate piecewise linear process $S^{(n)}$ with nodes at $D_n$, I find that $$ \forall \alpha>0\quad\forall \epsilon>0\quad \exists N\quad \forall n\geq N \quad \mathbb{P}(\sup_{[0,1]}|S^{(n)}_t-B_t|<\epsilon)>1-\alpha. $$ Hence (assuming $f$ piecewise linear with nodes at $D_n$) $$ \mathbb{P}(\sup_{[0,1]} |S^{(n)}_t-f(t)|<\epsilon)=:C(\epsilon,\alpha)>0. $$ Combining I get $$ \mathbb{P}(\sup_{[0,1]} |B_t-f(t)|<\epsilon)\geq C(\epsilon,\alpha)-\alpha, $$ which however might very well be negative depending on the behavior of $C(\epsilon,\alpha)$.
Any idea on how to fix things?