I have been brushing up on stochastic analysis before the start of my PhD, and I encountered this exercise on the book "Stochastic Calculus with applications" by Paolo Baldi. The text is as follows.
Let $B=(B_t)_{t \in \mathbb{R}_{+}}$ be a Brownian Motion. Prove that for every $a>0,T>0$,
$$P(B_t \leq a \sqrt{t} \ \ \forall \ t \in [0,T])=0$$
Now, my first instinct was to simplify the problem to get some intuition in the matter, and thus I tried to find
$$P(B_t < a \ \ \forall \ t \in [0,T])$$
What I did was applying the reflection principle, that is,
$$ \begin{align*} P(B_t < a \ \ \forall \ t \in [0,T])&=P(\max_{t \in[0,T]}B_t < a)=1-P(\max_{t \in[0,T]}B_t \geq a) \\ &= 1-2P(B_T\geq a)=P(B_T < a) - P(B_T \geq a) \end{align*}$$
Now, since $a$ is strictly positive, $P(B_T < a) > \frac{1}{2}$ and $P(B_T \geq a)<\frac{1}{2}$ and thus we obtain $P(B_t < a \ \ \forall \ t \in [0,T])>0$.
Now, while the derivation I have performed is not in line with what I was asked, I am a bit stuck. I do not understand how the result that was asked at the beginning can possibly hold, if my derivation is correct: intuitively, the $\sqrt{t}$ factor should at best strengthen my result. So I am starting to think that I simply made a mistake, but I cannot find it.
The issue with your approach is that it doesn't show a meaningful way to relate the first hitting time of the level $a$ and the first hitting time of the curve $a \sqrt{t}$, such that you will obtain your required bound.
Instead, consider the following: $$A = \left\lbrace B_t \leq a \sqrt{t} \text{ for all } t \in [0,T] \right\rbrace = \left\lbrace \frac{B_t}{\sqrt{2 t \log \log t^{-1}}} \leq \frac{a}{\sqrt{2 \log \log t^{-1}}} \text{ for all } t \in [0,T] \right\rbrace$$ By the local modulus of continuity for Brownian motion, almost surely we have $$\limsup_{t \to 0} \frac{B_t}{\sqrt{2 t \log \log t^{-1}}} = 1$$ yet, $$\limsup_{t \to 0}\frac{a}{\sqrt{2 \log \log t^{-1}}} = 0$$ Thus, $\mathbb{P}(A) = 0$.