Let $K\subseteq \mathbb R$ be non-empty and $f: K\to K$ be continuous such that $$|x-y|\leq |f(x)-f(y)|~\forall x,y\in K.$$ Which of the following statements are true?
$1.$$f$ need not surjective.
$2.$$f$ must be surjective if $K = [0,1]$.
$3.$$f$ is injective and $f^{-1}: f(K)\to K$ is continuous.
$4.$$f$ is injective, but $f^{-1}: f(K)\to K$ need not be continuous.
If $ K = \mathbb{R} $, it can be proved that $ f $ is both one-to-one and onto by showing that the limits of $ f $ as $ x \to \infty $ and $ x \to -\infty $ are $\infty$ and $-\infty$, and then using the intermediate value property to establish surjectivity.
However, how can we show that $ f $ is surjective when $ K = [0, 1] $? Proving injectivity is straightforward since $|x - y| > 0 $ implies $ |f(x) - f(y)| > 0 $?
For the last two options, there is a problem because $ K $ is not necessarily an interval, so we cannot directly apply the inverse function theorem to guarantee the continuity of $ f^{-1} $.
Please help me solve this problem, as the given answer is that statements 1, 2, and 3 are true.