Exhibiting a counter-example is straight-forward enough. For example, let $B_{t}(\omega)$ be a Brownian motion and $\mathcal{T}(\omega)$ a stopping time on $(\Omega,\mathbb{P})$ with a continuous distribution.
Then with
$$B'_{t}(\omega)=\left\{\begin{array}{ll}B_{t}(\omega),&t\neq\mathcal{T}(\omega)\\B_{t}(\omega)+1,&t=\mathcal{T}(\omega),\end{array}\right.$$
$B'_{t}(\omega)$ satisfies (1) and (2) below, but is discontinuous precisely when $t=T(\omega)$. Therefore, $B_{t}(\omega)$ is a particular realization of Brownian motion that is not everywhere continuous.
There are lots of other ways to obtain a "bad" Brownian motion. Another example is
$$B'_{t}(\omega)=B_{t}(\omega)\mathbb{1}_{\{B_{t}(\omega)\;\text{irrational}\}},$$
but this is less straight-forward to prove.
The reason for stipulating almost sure continuity has to do with the way one constructs Brownian motion, and the issue can be completely dispensed with dependent on one's approach.
The usual presentation in finance texts is the abstract one, namely given a probably space $(\Omega,\mathbb{P})$, one has a Brownian motion $B_{t}(\omega)$ on this space if
- For every set of times $0\leq t_{1}<t_{2}<\ldots<t_{n}$ the increments $B_{t_{1}},B_{t_{2}}-B_{t_{1}},\ldots,B_{t_{k}}-B_{t_{k-1}}$ form a mutually independent set of random variables on $(\Omega,\mathbb{P}).$
- The increments above are normally distributed with mean $0$ and variance $\Delta t$.
- For almost every $\omega\in\Omega$ the path $t\mapsto B_{t}(\omega)$ is continuous.
Most texts also include a section that sketches a concrete realization of Brownian motion as the limit of scaled random walks. If one does this rigorously, one sees that (3) upgrades to for every $\omega\in\Omega$
Indeed, if we start with $(\Omega,\mathbb{P})$ satisfying the above and let $\mathcal{P}$ denote the collection of continuous functions $[0,\infty)\to\mathbb{R}$ with $p(0)=0$, then we get from (3) the inclusion map
$$\mathcal{i}:\Omega\to\mathcal{P},$$
defined on a set $\Omega'\subset\Omega$ of full measure, and the push-forward measure of $\mathbb{P}$ onto $\mathcal{P}$ under this inclusion map turns out to be equal to the Wiener measure $\mathbb{W}$ on $\mathcal{P}$, which is unique.
Conversely, one can construct $(\mathcal{P},\mathbb{W})$ directly by starting with the set $\mathcal{P}$ (where every element of this set is continuous a priori) and demonstrating that the measures $\mu_{N}$ on $\mathbb{Z}^{\infty}_{2}$ arising from the appropriately scaled random walks $S_{t}^{N}(\omega)$ ($\omega\in\mathbb{Z}^{\infty}_{2})$ induce a collection of tight measures on $\mathcal{P}$ which converge weakly to $\mathbb{W}$:
$$\mu_{N}\Longrightarrow\mathbb{W}\;\text{(weakly)}$$
One then defines
$$\tilde{B}_{t}(\omega):=p(t)\in\mathcal{P}$$
and readily shows that under $\mathbb{W}$, $\tilde{B}_{t}$ satisfies (1)-(3) and that therefore
$$\tilde{B}_{t}(\omega)=B_{t}(\omega),$$
but that now every Brownian motion is continuous.
The equivalence of the implications above show the existence of Brownian motion is essentially tantamount to the existence of a Wiener measure on $\mathbb{W}$ arising from the sequence of measures arising naturally from the scaled random walks. If one starts from the goal of obtaining this measure, one gets continuity for every Brownian motion $p(t)=B_{t}(\omega)$.
Other constructions of Brownian motion require us stipulate almost sure continuity due to technicalities arising from measure theory on product spaces. The quickest construction of Brownian motion in this direction is by applying Kolmogorov's extension theorem on a suitable class of processes; details can be found in Durrett.