In page 128 of Ethier and Kurtz(1986 - Markov processes, convergence and characterization) one reads:
What is the converse here?
A stochastic process with sample paths in $D_{E}[0,\infty)$ is a $D_E[0,\infty)$- valued random variable.
If so then why is this not true when $E$ is not separable?
Let $(\Omega,\mathcal F,\Bbb P)$ be a probability space and $X:\Omega\times[0,\infty)\to E$ a mapping such that $t\mapsto X(\omega,t)$ is right-continuous on $[0,\infty)$ with left limits on $(0,\infty)$, for each $\omega\in\Omega$. In particular, $X(\omega,\cdot)$ is an element of $D_E[0,\infty)$ for each $\omega\in\Omega$. Let $\mathcal B$ denote the Borel $\sigma$-algebra on $D_E[0,\infty)$; here we think of $D_E[0,\infty)$ as a metric space (endowed with a metric compatible with the Skorokhod $J_1$ topology). If $E$ is separable, then so is $D_E[0,\infty)$. Then $X$ is a "$D_E[0,\infty)$-valued random variable" precisely when $X$ is $\mathcal F/\mathcal B$-measurable.
On the other hand, to say that $X$ is a "stochastic process" is to say that $X_t:\omega\to X(\omega,t)$, mapping $\Omega$ to $E$, is $\mathcal F/\mathcal E$-measurable for each $t\ge 0$. (Here $\mathcal E$ is the Borel $\sigma$-algebra on $E$.) By standard measure theory, this is the same as saying that $X$ is $\mathcal F/\mathcal D$-measurable, where $\mathcal D:=\sigma\{\pi_t:t\ge 0\}$ and $\pi_t:D_E[0,\infty)\to E$ is the $t$th coordinate projection on $D_E[0,\infty)$. That is, for $x\in D_E[0,\infty)$, $\pi_t(x) = x(t)$.
It is true in general that $\mathcal D\subset\mathcal B$ (whence the assertion made by Ethier and Kurtz); the reverse inclusion, hence the equality $\mathcal D=\mathcal B$ is true if $E$ is separable. In this case, a stochastic process with sample paths in $D_E[0,\infty)$ is a $D_E[0,\infty)$-valued random variable.
