Proving there is no continuous function $f: [0, 1] \rightarrow \mathbb R$ that is onto I know that onto means for every $y \in Y$ there is an $x \in X$ s.t. $f(x) = Y.$
In this case we are saying there is no continuous function that exist that where for every $y \in\mathbb R$ there is an $x\in [0,1]$ such that $f(x) = \mathbb R$
But how would I go about proving no such function exists.
 A: Ilham's answer is great.  More generally, you should prove that the continuous image of a compact set is compact.  
Is $[0, 1]$ compact?  What about $\mathbb{R}$?
A: What does the extreme value theorem say? Can a bounded function attain every value in $\mathbb{R}$?
A: You could use that the image of a compact set under a continuous function is compact (if you know what that is) hence bounded, while $\mathbb R$ is not bounded, or use the extreme value theorem (for a closed bounded interval). Or, you could try a direct proof along the following lines.  
Suppose there was such an onto function $f: [0, 1] \to \mathbb R$. Then for every positive integer $n$ there is $x_n\in[0,1]$ such that $f(x_n)=n$. Since the sequence $x_n$ is bounded, in the closed interval $[0,1]$, it mush have a convergent subsequence $x_{n_k}$, say 
$\lim\limits_{k\to\infty}x_{n_k}=x\in[0,1]$. (Use Bolzano-Weierstrass theorem.) By continuity $\lim\limits_{k\to\infty}f(x_{n_k})=f(x)$. On the other hand $\lim\limits_{k\to\infty}f(x_{n_k})=\lim\limits_{k\to\infty}n_k=\infty$, so we get $f(x)=\infty$, a contradiction. 
