Do the closed unit disk $D$ and $f(D)$ intersect, if $||f(x)-x||\le2$ for all $x\in D$?

Because $\Vert f(x)-x\Vert\leq2$, the function $x\mapsto \frac12(x-f(x))$ maps $D$ continuously into itself and therefore has a fixed point $a\in D$. Then $f(a)=-a$ is in $f(D)\cap D$.

This is true. Let's fix $n$ and let $S=\{x \in \mathbb R^n : \|x\| = 1\}$ and $D=\{x\in\mathbb R^n : \|x\| \leq 1\}$. Let's suppose that we had some map $f:D \rightarrow \mathbb R^n\setminus D$ such that $\|f(x)-x\| \leq 2$ everywhere. We make the following observation: if $x$ is on $S$, then $f(x)$ cannot be antipodal to $x$.

More precisely, define $C:\mathbb R^n\setminus D\rightarrow S$ by mapping $x$ to $\frac{x}{\|x\|}$ and let $f'$ be the restriction of $f$ to $S$. We observe that

$$C(f'(x)) \neq -x.$$

This implies that $C\circ f'$ is homotopic to the identity because we can, for each $x$, draw a great circle segment connecting $C(f'(x))$ to $x$ and then linearly interpolate upon that segment for each $x$ to create a homotopy from $C\circ f'$ to $\operatorname{id}_S$. However, this contradicts that $f'$ must be homotopic to a constant function because it is the boundary of a function on $D$, so no such $f$ exists.