Suppose I have a sequence of closed sets $\{A_n\}_{n=1}^{\infty}$ in $\mathbb{R}^n$ and all the $A_n$'s are nonempty. In addition, we have $lim_{n \to \infty}d(x,A_n)=\infty$, where $x\in \mathbb{R}^n$. I need to show $\cap_{n=1}^{\infty}A_n^c$ is open.
I am not sure how to use the condition that $lim_{n \to \infty}d(x,A_n)=\infty$. I tried proving $\cup_{n=1}^{\infty}A_n$ is closed by arriving at a contradiction by supposing there is a point that is a limit point of $\cup_{n=1}^{\infty}A_n$ but is not contained in $\cup_{n=1}^{\infty}A_n$, but did not really get anywhere.
Suppose $\{y_j\}$ is a sequence in $\cup_n A_n$ converging to a point $y$. For each $j$ choose $n_j$ such that $y_j \in A_{n_j}$. Claim: $\{n_j\}$ is bounded. For this note that $d(x,A_{n_j})\leq d(x,y_j)$. By hypothesis $n_j \to \infty$ implies $d(x,y_j) \to \infty$ and hence $d(x,y)=\infty$, a contradiction. If $\{n_j\}$ is unbounded we again get a contradiction by going to a subsequence which goes to $\infty$. Hence $\{n_j\}$ is bounded. This implies that some integer $m$ is repeated infinitely many times in $\{n_j\}$. This in turn shows that $y_j \in A_m$ for infinitely many $j$, so $y =\lim y_j \in A_m$. Hence $ y \in \cup_n A_n$ proving that $\cup_n A_n$ is closed.