Your mathematical sense problably twitched when you read the title, as a simple counterexample of the theorem is some one-to-one function. Where then, is the mistake in this proof?
Let $f:\mathbb{R}\to \mathbb{R}$ be arbitrary.
We will prove by induction that for every $n\ge 1$, $(|A| = n \wedge A\subseteq\mathbb{R})\to \exists c\in\mathbb{R}:\forall x\in A:(f(x) = c)$.
Base case: let $n=1$ and $|A| = 1$ and let $a$ be the only element in A. Then since $a\in\mathbb{R}$ by definition $\exists c\in\mathbb{R}:(f(a) = c)$ and thus $\exists c \in\mathbb{R}:\forall x\in A:(f(x) = c)$.
Induction step: Suppose $n\ge 1$ and $\forall A\in P(\mathbb{R}):((|A| = n)\to \exists c\in\mathbb{R}:\forall x\in A:(f(x) = c))$. Let $A\subseteq\mathbb{R}$ and $|A| = n+1$. Choose some arbitrary $a_1\in A$ and let $A_1 = A\setminus\{a_1\}$. Then $|A_1| = n$ and thus there is some $c_1 \in\mathbb{R}$ such that $\forall x\in A_1 :(f(x) = c_1)$. All that there is left to prove is that $f(a_1) = c_1$. Let $a_2\in A$ and $a_2\neq a_1$. As before this means that there is some $c_2 \in\mathbb{R}$ such that $\forall x \in A_2:(f(x) = c_2)$. Let $a_3\in A$ and $a_3\neq a_1$ and $a_3\neq a_2$. Then $a_3\in A_1$ and $a_3\in A_2$, so $f(a_3) = c_1 = c_2 = f(a_3)$ , so since $ a_1\in A_2,\ f(a_1) = c_1$.
If $A$ has exactly two elements there is no such $a_3$.