Sequentially compact set

Question

Let $X$ be a topological space. We say a set is sequentially compact if for every sequence of points in $X$ we can extract a subsequence that converges to a point in $X$. Why don't we define it as every sequence of points in $X$ converges to a point in $X$?

Answer 1

Yes, the part in red is exactly the error. To conclude that $(x_n)$ converges, you need to know that every subsequence has a subsubsequence that converges to the same point.

The condition that every sequence in $X$ converges to a point of $X$ is extremely strong. For instance, pick two points $a,b\in X$ and consider a sequence that alternates between $a$ and $b$. If the topology on $X$ is reasonably nice (e.g., $T_1$), then this sequence cannot converge. But every subsequence has a subsubsequence that converges, since there is always a subsubsequence that is constant (but it varies whether it is constantly $a$ or constantly $b$).