Let us start with a simple example. Assume that $( \Omega, \mathcal{F}, P)$ is some probability space and let $X : \Omega \rightarrow [0, 10]$ be a random variable which is uniformly distributed on $[0,10]$.
Let us define a stochastic process $X( t, \omega)$ in the following way:
$$
X ( t, \omega ) := t X(\omega), \quad ( t, \omega) \in [0, \infty) \times \Omega.
$$
Now let us first fix $t=100$. Then $X ( 100 , \omega) = 100X(\omega)$. In other words, we get $100X$, which is again a random variable, since a random variable multiplied by a constant is again a random variable. This argument can be used to conclude that $X(t, \omega) = t X(\omega)$ is a random variable for every fixed $t \geq 0$.
Let us now fix some $\omega_0 \in \Omega$ and assume that $X(\omega_0) = 3$. This then means that for all $t \geq 0$
$$
X( t, \omega_0 ) =tX(\omega_0) = 3t.
$$
And if we let $t$ vary over $[0, \infty)$ we will simply get a linear function of the form
$$
y(t)=t X( \omega_0) = 3t, \quad t \in [ 0, \infty)
$$
which will be the sample path of the stochastic process $X(t, \omega)$ for the fixed $\omega_0 \in \Omega$.
Extending this argument, we see that all the sample paths of the stochastic process $X(t,\omega) = t X( \omega )$ are from the following class of linear functions:
$$
\{ y : [0, \infty ) \rightarrow \mathbb{R} : y(t) = bt, \ b \in [ 0, 10 ]\}.
$$
Let us now turn to a Brownian $B(t, \omega)$ defined on (some other) probability space $( \Omega, \mathcal{F}, P )$. The Brownian motions starts almost surely at $0$, which means that $B(0, \omega)$, which is just a random variable for the fixed $t = 0$, is equal to $0$ with a probability of $1$, i.e. $P( B(0, \omega ) = 0 ) = 1$. Moreover, we know that a Brownian motion has continuous sample paths, which means that for every fixed $\omega_0 \in \Omega$, $y(t) = B(t, \omega_0)$ is a continuous function of $t$. Therefore, with probability $1$, the sample paths of the Brownian motion will be from the following class of functions
$$
\{ y : [0, \infty ) \rightarrow \mathbb{R} : y(t) \ \text{is continuous}, \ y(0) = 0 \}.
$$