Show that; $S\cup A$ is sequentially compact if A is sequentially compact and $\lim\operatorname{dist}(x_n,A)=0$ Let $(X,\rho)$ be a metric space. If $S\subset X$ we define $\operatorname{dist}(x,S):=\mathrm{inf}\{\rho(x,y):y\in S\}$. Suppose $A\subset X$ sequentially compact. $(x_n)\subset X$ sequence such that $\lim\operatorname{dist}(x_n,A)=0$. $$S:=\{x_n:n\in\mathbb{N}\}$$ Could you help me to prove that $S\cup A$ is sequentially compact?
 A: Let$$L=\{x\in X\,|\,x\text{ is the limit of a subsequence of }(x_n)_{n\in\mathbb N}\}.$$It follows from the fact that $\lim_{n\to\infty}\operatorname{dist}(x_nA)=0$ and from the fact that $A$ is closed (since it is compact) that $L\subset A$. Therefore$$S\cup A=S\cup L\cup A.$$Since both $S\cup L$ and $A$ are compact sets, $S\cup L\cup A$ is compact, that is $S\cup A$ is compact.
A: Let $z_n \in S\cup A$ be sequence
If $A$ contain infinite many term of A then  since A is sequentially compact there is subsequence, $z_{n_k}\in A\subset A\cup S$ converging in $A\subset A\cup S.$
If $A$ only contain finite term for $z_n$. Then $z_n$ is identical to infinte term of $x_n$. say, $x_{n_j} = z_{n_j}$ 
  since $$\lim_{n\to\infty}d(x_n,A ) = 0$$
we have that 
$$\lim_{j\to\infty}d(z_{n_j},A ) = \lim_{j\to\infty}d(x_{n_j},A )  =0$$
Then there exists $y_j\in A$ such that $$\lim_{j\to\infty}d(z_{n_j},y_j ) =0$$
But $A $sequencailly compact them  there exist $(y_{j_k})_k$ subsequence of $y_j$ converging to some $y\in A$
we have 
$$\lim_{k\to\infty}d(z_{n_{j_k}},y)  \le\lim_{k\to\infty}d(z_{n_{j_k}},y_{j_k})+\lim_{k\to\infty}d(y_{{j_k}},y) =0$$
That is $$z_{n_{j_k}}\to y\in A\subset A\cup S.$$
