You don't really need the implicit function theorem for this. It suffices to prove that there is no injective continuous map $f:\mathbb R^2\to\mathbb R$. To see this, define $A_x = f(\{x\}\times\mathbb R)$ for each $x\in\mathbb R$. Since $\{x\}\times\mathbb R$ is connected and $f$ is continuous, $A_x$ must also be connected. Therefore it is an interval. (Either open, closed or half-open.) It is not degenerate (a point), because $f$ is injective. Therefore each $A_x$ contains an open interval $(p_x,q_x)$, where $p_x$ and $q_x$ are distinct rational numbers. Since there are only countably many such intervals $(p_x,q_x)$ and uncountably many sets $A_x$, there exist $x\neq y$ such that $(p_x,q_x)=(p_y,q_y)$. Thus the sets $A_x$ and $A_y$ intersect, which contradicts the fact that $f$ is injective. The proof is complete.