The question is: If $\displaystyle g:\mathbb{R}^2\to \mathbb{R}^2$ is a continuous map and there is $\displaystyle r>0$ such that $\displaystyle g(x)\in B(x, r)$ (ball center $\displaystyle x$, radius r) for every $\displaystyle x\in \mathbb{R}^2$, then $\displaystyle g(\mathbb{R}^2)=\mathbb{R}^2$.

I have an idea of a proof but I'm not sure how to formalize it using liftings, etc (I'm not well-practiced in doing calculations with non-trivial fundamental groups). After supposing there is $\displaystyle a$ in $\displaystyle \mathbb{R}^2$ such that $\displaystyle g(\mathbb{R}^2)\subseteq \mathbb{R}^2-\{a\}$, my idea is to consider the standard loop in $\displaystyle \mathbb{R}^2$, $\displaystyle e^{2\pi it}, \, t\in[0,1]=:I$, but with radius $\displaystyle 2r$ and centered at $\displaystyle a$, call it $\displaystyle \alpha$. Each point $\displaystyle \alpha(t)$ has image under $\displaystyle g$ contained in $\displaystyle B(\alpha(t), r)$, and so $\displaystyle g(\alpha(I))$ is contained in the annulus, center a, with inner radius r, outer radius 3r. Put $\displaystyle \beta=g\circ \alpha$. Since $\displaystyle g$ is continuous and $\displaystyle \alpha$ is homotopic to the constant loop at $\displaystyle a$, $\displaystyle \beta$ should be homotopic to a constant loop. But joining points of $\displaystyle \beta$ in the balls around the four axes-intercepts $\displaystyle \alpha(0),\, \alpha(1/4),\, \alpha(1/2), \, \alpha(3/4), \, \alpha(1)$, for example, (and perhaps winding around again to be certain), we can see that $\displaystyle \beta$ has to wind around $\displaystyle a$. So since $\displaystyle \beta(I) \subseteq \mathbb{R}^2-\{a\}$, it isn't homotopic to a constant loop, a contradiction.

It's frustrating that I have the idea but I am not not familiar enough with performing calculations to formalize this. If someone could help me get started I'd be most appreciative!

Many thanks