Here is an outline.
Suppose NOT: .
Because K is sequentially compact has an accumulation point (limit point) is K. Call it k.
Because O covers K, .
Do you see the contradiction there?
Let K be a sequentially compact subset of R^n and suppose O is an open set of R^n that contains K. Prove that there is some positive number r such that for any point U in K, Ball (U) with radius of r is contained in O.
Hint: by contradiction.
Good job P.
Now here's another way to see this, using the properties of distance functions.
Define the distance function denoting the boundary of O (which is nonempty as O is open). This is continuous as the norm is continuous. If we show that , then for all radii and all we will have .
Suppose on the contrary . Then there is a sequence such that . Since K is sequentially compact, the sequence has a subsequence that converges to some element .
So we get . This means , as is closed; This contradicts the fact that .