Assume $f:\mathbb{R}^n\rightarrow[0,1]$ is an injective map. Take any circle $C\subset \mathbb{R}^n$. If $f$ is continuous on $C$, then the restriction of $f$ to $C$ is a homeomorphism $C\to f(C)$ (because $C$ is compact), which is a contradiction, because $C$ minus any point is connected, while $f(C)$ minus any point in $f(C)\cap(\inf f(C),\sup f(C))$ is disconnected. This shows that $f$ has a discontinuity on any circle in $\mathbb{R}^n$. In particular, when $n>1$, the cardinality of the discontinuities is continuum.

**Added:** By a similar argument combined with the Borsuk-Ulam theorem, the number of discontinuities of any injective map $f:\mathbb{R}^n\to\mathbb{R}^{n-1}$ is continuum.