When is the infimum of an arbitrary family of measurable functions also measurable?

If you want a (sometimes useful) positive result: suppose $f_i : \mathbb R \to \mathbb R$ are continuous. Then $g(x) := \inf_{i \in I} f_i(x)$ is upper semicontinuous, and therefore (Borel) measurable. More general domains are possible, of course.

As you expected, the infimum of continuum many measurable functions need not be measurable, even in the case where $X$ is the real line with Lebesgue measure. In fact, if $A$ is any subset of $\mathbb R$ (in particular not necessarily measurable), its characteristic function is the infimum of at most continuum many measurable functions, namely the characteristic functions of the sets $\mathbb R-\{x\}$ for all $x\notin A$.

A similar argument shows that even the infimum of $\aleph_1$ measurable functions won't be measurable if, in your measure space, the intersection of some $\aleph_1$ measurable sets is not measurable. And there are such measure spaces. For example, take $X$ to be a set of cardinality $\aleph_1$, take $\Sigma$ to be the $\sigma$-field of countable and co-countable sets, and take $\mu$ to give the countable sets measure 0 and the co-countable sets measure 1.

Whether the same thing happens for Lebesgue measure on $\mathbb R$ is independent of the usual ZFC axioms of set theory. It is consistent with ZFC that there is a non-Lebesgue-measurable set of cardinality $\aleph_1$, and it is also consistent that every union of $\aleph_1$ sets of Lebesgue measure 0 has measure 0, which implies that every union or intersection of $\aleph_1$ measurable sets is measurable.

The smallest cardinal $\kappa$ such that the union of some $\kappa$ sets of Lebesgue measure 0 does not have measure 0 is called the additivity of Lebesgue measure; it is one of the many well-studied cardinal characteristics of the continuum. It is also the smallest $\kappa$ such that the infimum of some $\kappa$ Lebesgue measurable functions $\mathbb R\to\mathbb R$ fails to be measurable.

Here is a set of of measurable functions with cardinality the continuum whose infimum is not (Borel) measurable:

Let $S\subset [0,1]$ be a non-measurable set. For $t\in[0,1]$ let $f_t(x)$ be the function defined as follows:

If $t \in S$ then $f_t(x) = 2$ for $t \not = x$ and $f_t(t) = 1$. If $t \not \in S$, then let $f_t(x) \equiv 2$. Then $f(x) := \inf_{t\in[0,1]} f_t(x)$ is $2$ on $S^c$ and $1$ on $S$, so is certainly not measurable.