Equivalent characterizations for Hausdorff in a first-countable space Today a student came to me with the following exercise from a course in topology:
Suppose $(X,\mathcal{T})$ is first-countable. Show that the following are equivalent:


*

*$X$ is Hausdorff

*Every convergent sequence in $X$ converges to precisely one point

*$X$ is T1 and every compact subset of $X$ is closed


I was able to show and explain all implications, except for one. 
I was not able to find a proof for $3. \Rightarrow 1$ or $3. \Rightarrow 2$. 
Could someone explain to me how I should handle this implication? 
Or could someone give a hint which characterization of Hausdorff is most appropriate for this proof?
As always, any help would be appreciated!
 A: To prove $3. \Rightarrow 2.$, consider a convergent sequence $(x_n)_{n\in\mathbb{N}}$, and let $x_\ast$ be any one of its limit points (we don't know yet that there is only one). The set
$$K = \{x_\ast\} \cup \{ x_n : n \in \mathbb{N}\}$$
is compact.
By $3.$, $K$ is closed. But any limit point of the sequence is contained in the closure of $\{ x_n : n \in \mathbb{N}\}$, so in particular in $K$.
For any $y \neq x_\ast$, consider a neighbourhood $U$ of $x_\ast$ that doesn't contain $y$ (that exists since the space is $T_1$), and the subsequence $(x_{n_k})$ of terms that lie in $U$ to conclude that $(x_n)$ doesn't converge to $y$: the subsequence still converges to all limit points of the original sequence, and the corresponding compact set $K' = \{x_\ast\} \cup \{ x_{n_k} : k \in \mathbb{N}\}$ by construction does not contain $y$; Since $K'$ is closed, $y$ has a neighbourhood that doesn't contain any of the $x_{n_k}$.
A: To see that $3. \Rightarrow 1.$, assume that $X$ is $T_1$ and all compact sets are closed. Suppose that $X$ is not Hausdorff, so there are two points $p,q \in X$ such that every neighbourhood of $p$ intersects every neighbourhood of $q$. 
As $X$ is first countable, $p$ and $q$ have local neighbourhood bases $U_n, n \in \mathbb{N}$ and $V_n, n \in \mathbb{N}$ resp. that we can choose decreasingly, without loss of generality. Pick points $x_n \in U_n \cap V_n$ by the non-Hausdorffness. Then $(x_n) \rightarrow p$ and $(x_n) \rightarrow q$, as sequences. (Note that argument already shows that $2. \Rightarrow 1.$, by contrapositive.) The set $\{x_n: n \in \mathbb{N}\} \cup \{p\}$ is compact (it's a convergent sequence) but not closed, as $q$ is in its closure but not in the set. So this would contradict $3.$, and so $X$ is Hausdorff.
As to $3. \Rightarrow 2.$: This last argument shows that if we have a sequence having two distinct limits, we have a compact set that is not closed (the sequence and one of the limits). 
