Showing second-countable + T$_1$ + compact implies sequentially compact 
Let $X$ be a second-countable T$_1$-space. If $X$ is compact then it is sequentially compact.

I know that compact implies countably compact. And under the condition that $X$ is first-countable: countably compact implies sequentially compact. Since $X$ is second-countable, thus first-countable, we have that $X$ is sequentially compact.
Is this correct? As the conditions of $X$ being a second-countable T$_1$-space weren't really used here, I suspect something is wrong.
However, those conditions may be needed to show that if $X$ is sequentially compact then it is compact.
 A: It is correct, and if one goes into a bit more detail, it is possible to eliminate the $T_1$ hypothesis.
Suppose that $X$ is compact and first countable, and let $\sigma=\langle x_n:n\in\Bbb N\rangle$ be a sequence in $X$. Let $A=\{x_n:n\in\Bbb N\}$. If $A$ is finite, $\sigma$ has a constant subsequence, which of course converges to a point of $X$. Otherwise, compactness of $X$ implies that the set $A$ has a limit point $p$.

We do not need $X$ to be $T_1$ for this to hold. If $A$ has no limit point, then each $x\in X$ has an open nbhd $U_x$ such that $U_x\cap A$ is finite. $\{U_x:x\in X\}$ is an open cover of $X$, so there is a finite $F\subseteq X$ such that $\{U_x:x\in F\}$ covers $X$. But then $A=\bigcup\{U_x\cap A:x\in F\}$ is a finite union of finite sets and so is finite, contrary to our assumption.

Now let $\{B_n:n\in\Bbb N\}$ be a countable local base at $p$; without loss of generality we may assume that $B_{n+1}\subseteq B_n$ for each $n\in\Bbb N$. (Why?) There is an $m_0\in\Bbb N$ such that $x_{m_0}\in B_0$. Given $m_k$ for some $k\in\Bbb N$, there is an $m_{k+1}\in\Bbb N$ such that $m_k<m_{k+1}$ and $x_{m_{k+1}}\in B_{k+1}$, since $B_{k+1}\cap A$ is infinite. In this way we can recursively construct a subsequence $\langle x_{m_k}:k\in\Bbb N\rangle$ of $\sigma$ such that $x_{m_k}\in B_\ell$ for all $k\ge \ell$, and clearly this subsequence converges to $p$. This shows that $X$ is sequentially compact.
