In a topological space, the latter two properties are obviously equivalent which can be seen by passing from a set to its complement. Sets that posses either are often called bicompact. In a metric space, all four properties are equivalent.
Every topological space can be embedded into a compact one, and this in a variety of ways. Let's consider, for example, the straight line R with the usual topology induced by the standard metric, d(x, y) = |x - y|. R is not compact because, for one, there is a sequence, say a sequence of positive integers, that contains no convergent subsequence. There is also a cover by open intervals (n - 1, n + 1), n = 0, 1, 2, ... that does not contain a finite subcover.
One way to embed R into a compact set is by adding two "end points" -∞ and +∞ and declaring their neighborhoods to be the sets {x: x < n} and {x: x > n}, respectively, where n is any real number. The resulting set will be very much like a finite closed interval, now [-∞, +∞] and will be compact because any cover will include (at least) two sets - one covering +∞, the other -∞ and the remaining cover a compact set (a closed finite portion of R.)
It's noteworthy, that in both variants of compactification the sequence of positive integers {n} becomes convergent, to boot, limn→∞n = ∞. For the two point compactification, limn→-∞n = -∞, whereas, for the one point compactification, limn→∞n = ∞, regardless, in a sense, how n moves to infinity - left or right. In particular, the sequence 1, -2, 3, -4, 5, ... converges to ∞.