Your intutive reasoning is leading you astray because you are thinking of Brownian motion as behaving like a smooth curve, for which there is a well-defined "direction" in which it is heading. Brownian motion isn't like that. (As you probably know, it's almost surely nowhere differentiable, so its "velocity vector" is undefined.)

Let $\tau$ be the time at which the Brownian motion first hits the sphere (note that $\tau$ is a stopping time). The probability that the particle "comes back inside", i.e. that there is a time interval $(\tau, \tau+\epsilon)$ (with $\epsilon$ allowed to depend on the sample space) during which the particle is inside the sphere, is 0. Likewise, the probability that the particle stays on the surface of the sphere for a positive time interval is also 0.

What actually happens after time $\tau$, with probability 1, is that the particle "vibrates" in such a way that during every time interval starting at $\tau$, no matter how short, it was sometimes inside and sometimes outside the sphere. Think about a function like $x \sin(\frac{1}{x})$ - it's continuous at 0, yet in every interval $(0, \epsilon)$ it takes both positive and negative values. So almost surely, the particle "immediately" exits the ball, and also "immediately" re-enters the ball. You can prove this by observing that the Blumenthal 0-1 law, combined with the strong Markov property, says that $\mathcal{F}_\tau^+$ is almost trivial - any phenomenon regarding what happens "immediately after" a stopping time must almost surely happen, or almost surely not happen.

So for your problem, first note that the probability that the sphere is hit *exactly* at time $t$ is 0: $P(\tau = t) = 0$ (indeed $\tau$ has a continuous distribution). So on the event that $\sup_{s \le t} \|B(s)\| \ge r$, the particle almost surely hit the sphere strictly before time $t$. This left it some positive interval of time before $t$, in which almost surely it exited the ball. Hence we have $\sup_{s \le t} \|B(s)\| > r$.