Does $T_1$ imply Fréchet–Urysohn (every limit point is a limit of some sequence)? It's not a problem from a book so I’m not even sure the statement is true. Nevertheless here's an alleged proof:
ADDED LATER: Although the result is wrong I can't find a problem with the proof. I would be thankful if someone would spot my mistake.
Let $X$ be a $T_1$ topological space and let $A \subseteq X$ be a subset with at least one limit point. let $x \in A$ be a limit point. Construct a sequence as follows:
let $U_1$ be a neighbourhood of $x$. $x$ is a limit point so $U_1$ contains a point in $A$ other than $x$, say $x_1$. $T_1$ is subspace hereditary and so $U_1$ is in itself $T_1$. By the $T_1$ property there exists neighbourhoods $V_1,U_2 \subseteq U_1$ of $x_1$ and $x$ respectively, such that $x \not\in V_1$  and $x_1 \not\in U_2$. 
By repeating the argument above every neighbourhood $U_n$ of $x$ contains a point $x_{n} \in A$ and every such point is contained in a neighbourhood $V_{n}$ with $x \not\in V_{n}$. Moreover there's a neighbourhood $U_{n+1}$ of $x$ with $x_{n+1} \not\in U_{n+1} \subset U_n$. By induction on $n$ we obtain the sequence of neighborhoods
$$\cdots  \subset U_3 \subset U_2 \subset U_1 \subseteq A $$ such that $\bigcap_{n=1}^{\infty}U_n=\{x\}$.
and a sequence of points  $\{x_i\}_{i=1}^{\infty}$ such that $x_i \in U_j$ iff $i \geq j$.
We will show that the sequence $x_n$ converges to $x$.  
Let $W$ be an arbitrary open set containing $x$. Since $\bigcap_{n=1}^{\infty}U_n=\{x\} \subset W$.  It must be that $U_N \subset W$ for some $N$. From this it follows that:
$$ n>N \implies x_n \in U_N \subseteq W $$
And thus $x_n$ converges to x.
Since I teach myself I have no one to check my exercises and so I’d really appreciate comments about my "writing style" whatever that may mean.
 A: No, absolutely not. The ordinal space $\omega_1+1$ consisting of the first uncountable ordinal and all of its predecessors with the order topology is a compact Hausdorff space that is not Fréchet-Uryson: the point $\omega_1$ is not the limit even of any countable set not containing it. Another compact Hausdorff counterexample is $\beta\Bbb N$, the Čech-Stone compactification of $\Bbb N$: no point of $\beta\Bbb N\setminus\Bbb N$ is the limit of a sequence.
Added: Getting the strictly decreasing sequence $\langle U_n:n\in\Bbb N\rangle$ of open nbhds of $x$ is fine, but you can’t legitimately conclude that $x$ is the only point in their intersection. When you do that, you’re actually assuming that the set $\{x\}$ is a $G_\delta$-set, i.e., a set that is the intersection of at most countably many open sets. The point $\omega_1$ in the space $\omega_1+1$ is a counterexample: the intersection of any such sequence of open nbhds of $\omega_1$ in that space is an uncountable set, hence certainly not just the singleton $\{\omega_1\}$.
There is actually a further problem beyond this. Even if $\bigcap_{n\in\Bbb Z^+}U_n=\{x\}$, it does not follow that $\langle x_n:n\in\Bbb Z^+\rangle$ converges to $x$. If $p\in\beta\Bbb N\setminus\Bbb N$, for instance, it’s possible to find open nbhds $U_n$ of $p$ and natural numbers $k_n\in\Bbb N$ such that $U_0\supsetneqq U_1\supsetneqq U_2\supsetneqq\ldots$, $\bigcap_{n\in\Bbb N}U_n=\{p\}$, $k_n\in U_n\setminus U_{n+1}$ for each $n\in\Bbb N$, and $\langle k_n:n\in\Bbb N\rangle$ does not converge to $p$ (because in fact no non-trivial sequence in $\beta\Bbb N$ converges to $p$).
