Subsequence implicate bounded and closed set I've been thinking about that problem for a long time, now it is right time to ask! Problem: > Proof that if $ K \subset \mathbb{R}^{d} $ is such a set that every sequence with elements in $ K $ include subsequence converge to element from $ K $, than $K$ is closed bounded and closed set Honestly I made few problems from that topic but I still dont have any idea how to do that proof. Please help! Greetings:)

If $K$ was not bounded then for every $n \in \mathbb{N}$ you had an $x_n \in K$ with $\vert x_n \vert > n$. Then the sequence $(x_n)_n$ would be diverging to $\infty$ and thus had no convergent subsequence. If $K$ was not closed then there would be an $x \in \bar{K}\setminus K$ and for every $n \in \mathbb{N}$ you had an $x_n \in K$ with $\vert x_n -x \vert < \frac 1 n$. Then the sequence $(x_n)_n$ would be converging to $x$ and so would every subsequence. But since $x \
otin K$ there would be no convergent subsequence with limit in $K$.