No, there are no such functions. Let $f,g$ be two functions from $\mathbb R$ to $\mathbb R$. Denote $\lbrace I_z=\left[\tfrac z 2, \tfrac {z+1} 2\right) \mid z \in \mathbb Z \rbrace$ the (countable) set of intervals that partition $\mathbb R$. Since $\mathbb R$ is uncountable, there is at least one $z_1 \in \mathbb Z$ such that $F=\{x \in \mathbb R \mid f(x) \in I_{z_1}\}$ is also uncountable. Similarly, since $F$ is uncountable, there is at least one $z_2 \in \mathbb Z$ such that $G=\{x \in F \mid g(x) \in I_{z_2}\}$ is also uncountable. Now, if $x$ and $y$ are two distinct numbers from $G$, then $|g(x)-g(y)|<\tfrac 1 2$ (as $g(x),g(y) \in I_{z_2}$ ) and $|f(x)-f(y)|<\tfrac 1 2$ (as $x,y \in F \implies f(x),f(y) \in I_{z_1}$ ), so $|f(x)-f(y)|+|g(x)-g(y)|<1$ .