Does the sequence lemma hold for Hausdorff? Munkres Topology: Sequence Lemma

Let $X$ be a topological space; let $A \subset X$. If there is a sequence of points of $A$ converging to $x$, then $x \in \overline{A}$; the converse holds if $X$ is metrizable. 

This lemma has been asked about before. Lemma 21.2 in Munkres' TOPOLOGY, 2nd ed: The Sequence Lemma
My question is about changing "metrizable" with "Hausdorff". I found a class where in a homework problem set, it says "This is an example of a non-Hausdorff space in which the sequence lemma holds." The saying says to me that the class has a version of the sequence lemma that holds for Hausdorff and then there's an example where the sequence lemma holds for something they would not expect, which is non-Hausdorff. (Another explanation is that the example is non-metrizable and then it is additionally pointed out that the example is also non-Hausdorff.) I hope that the version is exactly the same as the one above except for changing "metrizable" with "Hausdorff". (I cannot find anything of the sort in Week 4 or Notes for whole class.)
I have attempted a proof but it has some resemblance to the proof for metrizable spaces. I am choosing to not type up this proof fearing it is wrong because I have somehow assumed metrization.
 A: You need first countability for this to work. So the statement

If $X$ is Hausdorff, then $x\in \bar{A}$ implies that there is a sequence in $A$ converging to $X$.

is not true.
For example, consider $X=\omega_{1}+1=\{0,1,\ldots,\omega_{0},\omega_{0}+1,\ldots,\omega_{1}\}$ with the order topology (where $\omega_{1}$ is the union of all countable ordinals, i.e. the first uncountable ordinal). This is a Hausdorff space. The closure of $A=\{ x\in \omega_{1}+1 \mid x<\omega_{1} \}$ is then $X$, so $\omega_{1}\in \bar{A}$. But we cannot find a sequence in $A$ converging to $\omega_{1}$ (for cardinality reasons).
A: The "sequence lemma" ($x \in \overline{A}$ iff there is a sequence from $A$ converging to $x$) holds exactly in the class of sequential spaces. All first countable spaces are sequential (so all metrisable ones are).
A sequential space can be Hausdorff but need not be. These are independent of each other: the line with two origins is not Hausdorff and sequential. The co-countable topology on the reals is not Hausdorff and not sequential. Any metric space is Hausdorff and sequential and $\omega_1+1$ (or $\overline{S}_\Omega$) is not sequential but Hausdorff.
