About sequential compactness If $A\subset \mathbb R^n$ is compact, $x\in A$, and let $\{x_i\}$ be a sequence in $A$ such that every convergent subsequence of $\{x_i\}$ converge to $x$. Prove that $\{x_i\}$ also converge. 
How to even approach this problem, by contradiction?
 A: Assume that $(x_n)$ does not converge to $x$, which means that there exists $\epsilon>0$ such that for all $N$ there exists $n\geq N$ with $|x_n-x|\geq \epsilon$.
By a proper induction, this is in turn equivalent to the existence of a subsequence $(x_{n_k})$ such that $|x_{n_k}-x|\geq \epsilon$ for all $k$.
Now by compactness, there exists a subsequence $(x_{n_{k_m}})$ of the latter which converges. By assumption, it converges to $x$.
Since $|x_{n_{k_m}}-x|\geq \epsilon$ for all $m$, we can pass to the limit as $m$ tends to $+\infty$ and find
$$
0=|x-x|\geq \epsilon
$$
a contradiction.
So $(x_n)$ converges to $x$.
A: Hint: 
1) Show that every subsequence of $(x_i)$ has a further subsequence that converges to $x$. 
2) Show that the condition in 1) implies that $(x_i)$ converges to $x$ (the contrapositive is easy to prove).
A: Given $\epsilon>0$, we have to show that almost all $x_i$ are in $B(x,\epsilon)$.
We are given that for $y\ne x$ there is no subsequence converging to $y$, hence for some $r(y)>0$ there are only finitely many $x_i\in U_y:=B(y,r(y))$.
Then
$$ A\subseteq B(x,\epsilon)\cup\bigcup_{y\in A\setminus\{x\}}U_y$$
is an open cover of $A$. By compactness, there is a finite subcover, hence
$$ A\subseteq B(x,\epsilon)\cup \bigcup_{i=1}^nU_{y_i}.$$
Each $U_{y_i}$ contains only sinitely many members of the sequence, hence $B(x,\epsilon)$ must contain almost all members of the sequence, as was to be shown.
