The answer is Yes if $X$ is finite (and possibly also Yes if $X$ is infinite and has $F(x)$ is finite for all $x\in X$, but I don't know off the top of my head about that). The reason why the answer is positive is that this is a reformulation of Hall's Marriage Theorem.

For infinite $X$ consider the following example. Let $X = \omega$ and set

- $F(0) = \omega \setminus \{0\}$, and
- $F(n) = \{n\}$ for $n \geq 1$.

Then it is clear that an *injective* map $f: \omega\to \omega$ with $f(m) \in F(m)$ for all $m \in \omega$ can't exist: Suppose we have some map $f: \omega\to \omega$ with $f(m) \in F(m)$ for all $m$. Then $f(0) \in \omega\setminus\{0\}$, say $f(0) = p$. But the definition of $F$ forces $f(p) = p$. Therefore $0\neq p$ but $f(0) = f(p)$ so $f$ is not injective.