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If $f:\mathbb R^n \to\mathbb R^n$ is continuous and such that for each $x$ in the domain $\|f(x)-x\|\leq1$ holds, how can I prove that then $f$ must be onto?

As a hint, I have been told, for an $a\in\mathbb R^n$ look for a fixed point of the function

$$g_a(x):=x+a-f(x+a)$$

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2nd hint: Suppose that the value $a$ is not attained.

What can you say about fixed points of $g_a$ ?

And what does Brouwer give you when you look at the map of the closed unit ball under $g_a$?

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    I am generally not familiar with usage of Brauwer theorem, I only know it's statement. But by attained, do you mean such that there is no $x\in\mathbb R^n$ such that $f(x)=a$ – 2017-02-11
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    yes. Then try to apply Brouwer on $g_a$ for that value of $a$ on the closed unit ball and get a contradiction. Alternatively, simply take any $a$ and apply Brouwer directly on $g_a$ to see that there is a fixed point so $a$ is attained. – 2017-02-11