Let $(\Omega ,\mathcal F,\mathbb P)=([0,1],\mathcal B([0,1],m)$ where $m$ is the Lebesgue measure. Let $X_t=0$ for all $t$ and $Y_t(\omega )=\mathbb 1_{t=\omega }$.
Even if $X_t(\omega )=Y_t(\omega )$ for all $t$ and all $\omega \neq t$ (i.e. for all $t$, except at one point, they are the same processes), we have that $$\mathbb P(\forall t\geq 0, X_t=Y_t)=0.$$ So it looks that to be indistinguishable, we should have $X_t(\omega )=Y_t(\omega )$ for all $t$ and all $\omega $. But, if this would be the case, I guess that we would define indstinguishable processes as $X_t(\omega )=Y_t(\omega )$ for all $t$ and all $\omega $, and not as $\mathbb P(\forall t, X_t=Y_t)=1$. So, are there indistinguishable processes $(X_t)$ and $(Y_t)$ s.t. $X_t(\omega )= Y_t(\omega )$ for all $t$ and all $\omega $ doesn't hold ?
Two processes $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ are a modification of each other if
$$\mathbb{P}(X_t=Y_t)=1 \quad \text{for all $t \geq 0$},$$
i.e. for each $t \geq 0$ there exists an exceptional null set $N=N(t)$ such that $$X_t(\omega) = Y_t(\omega) \quad \text{for all $\omega \in \Omega \backslash N(t)$}.$$
If the processes $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ are indistinguishable, then the exceptional null set can be chosen independently from $t$, i.e. there exists a null set $N$ such that
$$X_t(\omega) = Y_t(\omega) \quad \text{for all $t \geq 0$, $\omega \in \Omega \setminus N$}.$$
Equivalently, $\mathbb{P}(\forall t \geq 0: X_t=Y_t)=1$. This means that the processes can differ on a null set. Note that null sets might be quite "large", depending on the measure which we consider.
Take, for instance, $\Omega = [0,1]$ with the Dirac measure $\mathbb{P}:=\delta_0$, and define
$$X_t(\omega) := \sin(\omega t) \qquad Y_t(\omega)=0.$$
We have $X_t(0)=0=Y_t(0)$ and so
$$\mathbb{P}(\forall t \geq 0: X_t=Y_t) \geq \delta_0(\{0\})=1.$$
Consequently, $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ are indistinguishable although they are (from the "natural" point of view) quite different processes. In particular, $X_t(\omega) \neq Y_t(\omega)$ for "many" $t$ and $\omega$.