It’s true in any topological space $X$ that a sequence $\sigma$ converges if and only if every subsequence of $\sigma$ converges. One direction is trivial: if every subsequence of $\sigma$ converges, then obviously $\sigma$ converges, since it’s a subsequence of itself. To prove the other direction, suppose that $\sigma=\langle x_n:n\in\omega\rangle$ converges to $x$, and let $\langle x_{n_k}:k\in\omega\rangle$ be any subsequence of $\sigma$. Let $U$ be any open nbhd of $x$. Then there is an $m\in\omega$ such that $x_n\in U$ whenever $n\ge m$. But $n_k\ge k$ for each $k\in\omega$, so $n_k\ge m$ whenever $k\ge m$, and therefore $x_{n_k}\in U$ whenever $k\ge m$. It follows that $\langle x_{n_k}:k\in\omega\rangle$ converges to $x$.