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I constructed a bijection by using decimal expansions of two real numbers and taking numbers 1 by 1 consecutively. (It took me hours to come up with this). I remember someone saying appealing to some sort of expansion is the only way to do this, and I think this type of method can never be continuous. Is there anyway to prove that no such function exists?

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I don't know, but continuous is probably too weak as a condition. In fact there is the Peano curve, which is an example of continuous surjection $[0,1] \to [0,1] \times [0,1]$. –  Andrea Apr 10 '13 at 12:23
This doesn't answer your question, but... "I constructed a bijection by using decimal expansions of two real numbers and taking numbers 1 by 1 consecutively." -- this is reminiscent of a trap I fell down when trying to solve this problem for the first time. Make sure you didn't assume that $\mathbb{R}$ is countable or anything like that! –  Clive Newstead Apr 10 '13 at 12:23
To expand on Willie Wong's link, the Peano curve $[0,1]\rightarrow [0,1]\times[0,1]$ can easily be extended to a continuous surjection $(0,1)\rightarrow\mathbb{R}^2$. Or in your case, by a composition by homeomorphism on the domain, a continuous surjection $\mathbb{R}\rightarrow\mathbb{R}^2$. –  Daniel Rust Apr 10 '13 at 13:11
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Yes, there exists a continuous surjection from $\mathbb R$ to $\mathbb R^2$. The following is a simple way to construct one, although there should be more elegant constructions. Let $f:[0,1]\to[-1,1]^2$ be a spacefilling curve which starts and ends at the origin, for example the Sierpinski curve (although one could of course start with any space-filling curve $[0,1]\to[0,1]^2$, obtain a new space-filling curve $[0,1]\to[-1,1]^2$ by translation and scaling, and then append paths to the beginning and end of the curve to make it start and end at the origin). By scaling, we can define a space-filling curve $g_n:[0,1]\to[-n,n]^2$ by $g_n(x)=nf(x)$ for any integer $n>0$, which starts and ends at the origin. We can now define a space-filling curve from $[0,\infty)$ to $\mathbb R^2$ by first walking along the curve $g_1$, then walking along the curve $g_2$, then walking along the curve $g_3$, and so on. This is a continuous curve and it passes through every point of $(x,y)\in \mathbb R^2$ since for every $(x,y)$ there exists an $n>0$ such that $(x,y)\in[-n,n]^2$.

To be explicit, the curve $F:\mathbb R\to\mathbb R^2$ defined by $F(x)=0$ if $x<1$, and $F(x)=g_n(x-n)$ if $x\in[n,n+1)$, is a continuous surjection.

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