Is it possible to construct a continuous and bijective map from $mathbb{R}^n$ to $[0,1]$?

The proof suggested by David Hartley here also works in this case, and I think it's worth noting here.

This proof is significantly cooler than the previous ones, because it proves something much stronger: If $X$ is a path-connected topological space admitting a continuous bijection $f:Xto [0,1]$, then $f$ is a homeomorphism. In particular, $X$ cannot be $Bbb{R}^n$, because $Bbb{R}^n$ is not homeomorphic to $[0,1]$.

Proof. Suppose $X$ is a path-connected space and that $f:Xto [0,1]$ is a continuous bijection. Let $phi:[0,1]to X$ be a path from $f^{-1}(0)$ to $f^{-1}(1)$. Composing $phi$ with $f$, we get a path from $0$ to $1$ in $[0,1]$; therefore, $fcircphi:[0,1]to [0,1]$ is surjective, and hence so is $phi:[0,1]to X$. Therefore $X$ is compact. Since $[0,1]$ is Hausdorff, this means $f$ is a closed map, and hence a homeomorphism.

To complete the answer of Chris Culter: this is also impossible for $n=1$. Suppose $f:Bbb{R}to [0,1]$ is continuous and bijective, and let $xin Bbb{R}$ be such that $f(x)=0$. Consider $f([x,infty))$; since $f$ is continuous, this is a connected subset of $[0,1]$ which contains $0$, and so there is some $r_1>0$ such that $[0,r_1]subseteq f([x,infty)$. Similarly, there is $r_2>0$ such that $[0,r_2]subseteq f((infty,x])$. Take $r>0$ such that $r<min{r_1,r_2}$. Then $r$ is both in $f((infty,x])$ and in $f([x,infty)$. Since $f$ is bijective, we know $f(x)neq r$, so we arrive at a contradiction: $r$ is both the image under $f$ of some number larger than $x$ and some number smaller than $x$.

EDIT. In fact, now that I think about it, the same proof works for every $n$ - just take, instead of $[x,infty)$ and $(infty,x]$, any two connected subsets $A,B$ of $Bbb{R}^n$ which both contain $x$ and such that $Acap B={x}$.

To answer the question in the title: No for $n>1$.

If $f:mathbb R^nto[0,1]$ is continuous and surjective then $f^{-1}([0,frac12))$ is a proper clopen subset of $f^{-1}([0,1]setminusfrac12)$. That means $f^{-1}([0,1]setminusfrac12)$ is disconnected. But $mathbb R^n$ minus a single point is connected, so $f$ must not be injective.