Suppose a set $K\subset \mathbb{R}$. Every sequence in K has a convergent subsequence that converges to a limit in K. Prove that K is bounded.
I want to use contradiction. Suppose K is not bounded, then there should be an unbounded sequence $\{a_n\}$. And I want to show that $\{a_n\}$ doesn't have a convergent subsequence, but I don't know how to do.
On
Suppose $\{a_n\}$ is a sequence that satisfies the following: for any $M$ there exists $N_M$ such that $|a_n| \ge M$ for all $n \ge N_M$. Such a sequence exists in $K$ because $K$ is unbounded. Moreover, any subsequence of $a_n$ satisfies this same property.
Suppose for sake of contradiction that $\{a_n\}$ has a convergence subsequence $\{a_{n_k}\}$ with some point of convergence $a$. Then for some $K$, we have $|a_{n_k}-a| < 1$ for all $k \ge K$, so $|a_{n_k}| < |a|+1$ for all $k \ge K$. This contradicts the fact that $\{a_{n_k}\}$ is also an unbounded sequence.
Clearly this won't work for any unbounded sequence: for example, the sequence $0,1,0,2,0,3,0,4,\ldots$ has a convergent subsequence. However, since $K$ is unbounded, we may choose $a_i$ to be any element of $K$ such that $|a_i| \geq i$; this way, we get a sequence $(a_n)$ with no convergent subsequence.