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

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

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

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 $x\in \Bbb{R}$ 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 $A\cap B=\{x\}$.