Problem 1.
… sequence $(x_n)$ in S there is a subsequence $(x_{n_k})$ which converges to some $x\in S$.
Problem 2.
… a sequence $(x_n)$ in S for which no subsequence converges to a point in S.
Problem.3.
A subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded (the Heine-Borel Theorem).
Problem 4.
If S is a single point x, there is only one sequence contained in S, the sequence x, x, … which converges itself.
Problem 5.
This fact is implied by the fact that the subsequence converges. In particular, if the subsequence is $(x_{n_k})$, then by definition of convergence for any $\epsilon>0$ every term with $k>N$ for some N is in the $\epsilon$-neighborhood of x. Since there are an infinite number of these, and they are all elements of the original sequence $(x_n)$, the statement must be true. $\blacksquare$
Problem 6.
We must show how to construct a subsequence $x_{n_1},x_{n_2},x_{n_3},\ldots$ which converges to x. Let $x_{n_1}$ be any element within the 1-neighborhood of x (that is, anything within distance 1 of the limit). Let $x_{n_2}$ be any element with $n_2>n_1$ (that is, some element of the sequence further out than $n_1$) and within the $\frac12$-neighborhood of x. Continue in this manner to select $x_{n_i}$ with $n_i>n_{i-1}$ in the $\frac1i$-neighborhood of x. Since there are an infinite number of points within each neighborhood of x to choose from, we can always find such points.
How can we prove that the subsequence $x_{n_k}$ converges to x? It suffices to show that every $\epsilon$-neighborhood of x contains some "tail" of the sequence, that is all $x_{n_k}$ with $k>N$ for some N. Choose N such that $0<\frac1N<\epsilon$. By construction if $k>N$ then $d(x_{n_k},x)<\frac1N<\epsilon$ since everything past $k=N$ is in a $\frac1{N+1}$-neighborhood or smaller. $\blacksquare$.
