Let $f:\mathbb{R} \to \mathbb{R}$ be a differentiable function.
Let's say that we want to apply the MVT on the interval $\left[0, \frac{1}{x}\right]$ , $\forall x\in (0,1]$.
Then we know that $\exists\, c_x \in \left(0, \frac{1}{x}\right)$ such that $f\left(\frac{1}{x}\right)-f(0)=\frac{1}{x}\cdot f'(c_x),\, \forall x\in (0,1)$.
What I would like to know is whether this $c_x$ is a function or not. I believe it is not, since there may be multiple $c_x$'s that work for some $x$, but many times I have encountered that in this case we would have $\lim\limits_{x\to\infty}c_x=0$ from the squeeze theorem.
What is the explanation? Furthermore, would it be better to call it $c(x)$ instead of $c_x$?
They're not saying that $c$ is unique for each $x$, simply that for each $x$, we can choose a $c_x$ that works. That there is such a function follows from an axiom of set theory, the axiom of choice.
You can call it $c(x)$ if you like, though I think $c_x$ is more common, but it doesn't really matter one way or the other.