We consider the space $\mathcal K(X)=\{K\subset X\colon K\mathrm{\ compact\ and\ non\ empty}\}$ endowed with the Hausdorff distance $\delta$ . Since Hausdorff metric inherits completeness, being $X$ complete, $(\mathcal K(X),\delta)$ is complete too. We then consider the mapping $T\colon \mathcal K(X) \to \mathcal K(X)$ defined by$$ T(A) = \bigcup_{i=1}^N T_i(A).$$ We claim that $T$ is a contraction. In fact, recalling that $\delta(A_1\cup A_2, B_1\cup B_2) \le \max\{\delta(A_1,B_1),\delta(A_2,B_2)\}$ while $\delta(T_i(A),T_i(B))\le \lambda_i \delta(A,B)$ if $T_i$ is $\lambda_i$ -Lipschitz, we have

So $T$ is a contraction on the complete metric space $\mathcal K(X)$ and hence, by Banach Fixed Point Theorem, there exists one and only one $K\in\mathcal K(X)$ such that $T(K)=K$ .