The following answer is not correct, as pointed out in the comments

$\textbf{Claim}$: Let $A \subset \mathbb{R}$ of continuum size and let $f: A \to \mathbb{R}$ then there exist $\mathcal{c}$ (contiuum size) many subsets of $A$ of size $\mathcal{c}$ such that for each such $X$, $f(X) \neq \mathbb{R}$.

$\textit{proof}$: Consider a partition $P$ of $A$ of size $\mathcal{c}$ such that each element of $P$ has size $\mathcal{c}$. If for $\mathcal{c}$-many elements of $P$ $f(X) \neq \mathbb{R}$ then we are finished thus we may assume w.l.o.g that for each $X \in P$ $f(X) = \mathbb{R}$. Then we can pick for each $r \in A$ and each $X_i \in P$ an $x_i \in X_i$ such that $f(x_i) = r$. Thus we obtain for each $r \in$ $A$ a set $Y_r \subset \mathbb{R}$ of size $\mathcal{c}$ such that $f(Y_r) \neq \mathbb{R}$.

Now let $f_1, f_2,..$ be an arbitrary sequence of functions from $\mathbb{R}$ to $\mathbb{R}$. By our claim there is a set $A_1$ of continuum size such that $f(A_1) \neq \mathbb{R}$. Then using our claim again there exists an $A_2 \subset A_1$ such that $f_2(A_2) \neq \mathbb{R}$, and an $A_3 \subset A_2$ and so on. 

One obtains a $\omega$-descending sequence $A_i$ of sets of reals of size continuum, thus its limit cant be empty as the continuum has uncountable cofinality. This gives us an infinte (though maybe not continuum size) set of reals $A$ such that for each $f_i$ $f_i(A) \neq \mathbb{R}$.