The sample paths of a measurable stochastic process are Borel

Let $(\Omega,\mathcal{F},P)$ be a probability space. Let $(X(t))_{t\geq 0}$ be a stochastic process (so the time domain is $[0,\infty)$).

Suppose that the stochastic process considered as a map $[0,\infty)\times\Omega\rightarrow\mathbb{R}$ given by $(t,\omega)\mapsto X(t,\omega)$ is measurable with respect to the product sigma-algebra $\mathcal{B}([0,\infty))\otimes\mathcal{F}$. Why then does it follow that for fixed $\omega\in\Omega$, the function $t\mapsto X(t)$ is a Borel function?

I'm thinking that the restriction of a measurable function to a measurable set is measurable, but the point set $\{\omega\}$ need not be measurable.