proving that a function from $\Bbb R^2$ to $\Bbb R^2$ is surjective

Define $g:B(0,2017)\rightarrow B(0,2017)$ by $g(y)=y-f(y)$. Obviously $g$ is continuous and it's image is $B(0,2017)$. By Brouwer's fixed point theorem link there exists $y_0\in B(0,2017)$ so that $g(y_0)=y_0$ in particular it follows that $f(y_0)=0$.

We are going to modify this argument a little bit and we will get that $f$ is onto

Let $x$ be any element in $\mathbb{R}^2$ and consider $g(y)=y+x-f(y+x)$ again, $g:B(0,2017)\rightarrow B(0,2017)$ and again there exists $y_0$ so that $g(y_0)=y_0+x-f(y_0+x)=y_0$ in particular $f(y_0+x)=x$ for every $x\in\mathbb{R}^ 2$

If $\|x\|\to\infty$ then clearly $\|f(x)\|\to\infty$. This means that after a $1$-point compactification of $R^2$, you can continuously extent $f$ to a function $S^2\to S^2$ via $f(\infty)=\infty$. Further, $f: S^2\to S^2$ is homotopic to the identity map, because the homotopy $(x,t)\mapsto x+t(f(x)-x)$ can be continuously extended to a pointed homotopy of the sphere via $(\infty,t)\mapsto\infty$. Hence the degree of $f$ is $1$ and $f$ is surjective.