Let $f_1,\dots,f_n:R^d\rightarrow R^d$ be diffeomorphisms. $\exists 0<c<1$ s.t. $\parallel Df_i(x)\parallel\leq c,\forall x\in R^d,\forall i$. Where the norm is the operation norm in $L(R^d)$.
How to proof that there exists a unique non-empty compact subset $K\subset R^d$ s.t. $K=\cup_{i=1}^nf_i(K)$
Let me flesh out the suggestion made in the comments to use Hausdorff distances. Let $C(\mathbb{R}^d)$ be the set of compact subsets of $\mathbb{R}^d$. Define $F: C(\mathbb{R}^d) \to C(\mathbb{R}^d)$ by $F(x) = \bigcup_{i} f_{i}(x)$.
Let's show that $F$ is a contraction on $C(\mathbb{R}^d)$. Let $X$ and $Y$ be two compact subsets of $\mathbb{R}^d$ with $d_{H}(X,Y) = \epsilon$, and let $M \subset X \times Y$ be a matching realizing this distance, i.e. $\forall (x,y) \in M$, $\|x - y\| \leq \epsilon$. Note that the Hausdorff distance, which is usually a infimum, can be taken to be a minimum for compact sets, although this isn't strictly necessary to make the proof work, it just makes it a little cleaner. Consider now $F(X)$ and $F(Y)$. Taking $x' \in F(X)$, we know $x' = f_{i}(x)$ for some $x \in X$ and some $i$. Let $y$ be paired with $x$ in $M \subset X \times Y$. Then $y' = f_{i}(y)$ is in $F(Y)$, and $\|x' - y'\| \leq c \|x - y\|$. A symmetric argument for an arbitrary element $y' \in F(Y)$ shows how to build a matching $M' \subset F(X) \times F(Y)$ realizing a Hausdorff distance of at most $c\epsilon$.
Applying the contraction mapping theorem shows that there is a unique fixed point for $F$, which is the set $K$ in question.