For any topological space $X$, let $\text{End}(X)$ denote the set of continuous functions $f:X\to X$. We say that ${\cal C}\subseteq \text{End}(X)$ *covers* $\text{End}(X)$ if for every $f\in \text{End}(X)$ there is $h\in {\cal C}$ and $x\in X$ such that $f(x) = h(x)$. 

I am interested in *minimal* covers (that is, covers that have the property that when you remove a member, it is no longer a cover). A boring example is the set of all constant functions which always has the same cardinality as $X$.

Note that $\text{End}(\mathbb{R})$ does have a countable cover: For $z\in\mathbb{Z}$ let $f_z:\mathbb{R}\to\mathbb{R}$ be defined by $r \mapsto r+z$. Let $${\cal C} = \{f_z: z\in\mathbb{Z}\}\cup\{{\bf 0}\}, $$ where ${\bf 0}$ denotes the constant zero function. Clearly this cover does not have a minimal subcover.

**Question.** Does ${\mathbb R}$ have a countable minimal cover?