# A problem on metric spaces and the distance of a sequence from a sequentially compact set

$(X,\rho)$ metric space. If $S\subset X$ we define $\mathrm{dist}(x,S):=\mathrm{inf}\{\rho(x,y):y\in S\}$. Suppose $A\subset X$ sequentially compact. $(x_n)\subset X$ sequence such that lim dist$(x_n,A)=0$. $S:=\{x_n:n\in\mathbb{N}\}$. Could you help me to prove that $S\cup A$ is sequentially compact?

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If $(y_n)$ is a sequence in $S\cup A$, then $(y_n)$ has a subsequence in $A$ (a case I'll leave to you) or a subsequence in $S$. If there is a subsequence in $S$, then, unless there is an element of $S$ that appears infinitely many times as a term (a case I'll leave to you), there is a subsequence of $(y_n)$ that is also a subsequence of $(x_n)$. If $(z_n)$ is a subsequence of $(x_n)$, then there is a sequence $(a_n)$ in $A$ such that $\lim_n\rho(z_n,a_n)=0$. There is a subsequence $(a_{n_j})$ of $(a_n)$ converging to some $a\in A$, and the subsequence $(z_{n_j})$ of $(z_n)$ with the same indices also converges to $a$.