Must a finer topology make fewer sequences converge? I think the question above is equivalent to this:

Let $(X,O)$ be a topological space, and let $A$ be a collection of subsets of $X$. If for any sequence $(x_n)$ that converges to any $x$ in $A$, there is an $N$ such that for all $n \ge N$, $x_n$ is in $A$, must $A$ be open?

My strategy for proving this has been: for each $x$ in $A$ find a $U_x$ which is open, contains $x$ and is a subset of $A$. Then $A = \cup_{x\in A} U_x$, which is open.
However, I can't show that such a $U_x$ must exist.
Thanks for any advice.
 A: Yes.  If $\tau$ is finer than T, then every $\tau$ convergent sequence is T convergent.  On the other hand tau convergent sequence may not be T convergent.  For example, when tau is discrete topology (the finest), the only convergent sequences are eventually constant sequences.  When T is indiscrete topology (the coarsest), every sequence converges to every point - a grand free for all.
A: To answer your first suspicion: indeed sequentially open sets are open, at least sometimes. Here a set $A$ is sequentially open iff
$$\forall x \in A: \forall (x_n) \text{ in } X: (x_n \to x) \rightarrow (\exists N: \forall n \ge N: x_n \in A) $$
Fact: in a metric space any sequentially open set is open. 
Proof: suppose $A$ is sequentially open, but not open. Then not open implies there is some $x \in A$ that is not an interior point of $A$. This implies then that for all $n \in \mathbb{N}$, $B(x, \frac{1}{n})\nsubseteq A$, so we can pick for each $n$ some $x_n \in B(x, \frac{1}{n})$ such that $x_n \notin A$. But by virtue of $x_n \in B(x, \frac{1}{n})$ we know that $x_n \rightarrow x$ (for let $\varepsilon >0$, pick $N$ so large that $\frac{1}{N} < \varepsilon$, then for $n \ge N$: $d(x_n, x) < \frac{1}{n} \le \frac{1}{N} < \varepsilon$, as required for convergence).
But then this sequence $(x_n)$ shows that $x$ does not belong in $A$, as $A$ wouldn't be sequentially open, as witnessed by $x$ and this sequence. So $A$ is open, as required.
Note that a minor adaptation of this proof shows that in any first countable space, sequentially open sets are open (the converse always holds, in any space). In fact this has inspired the definition of a whole new class of spaces: $X$ is called sequential whenever all sequentially open sets are open. In those spaces the topology is "determined by its convergent sequences". Not all spaces are sequential, though, as the example below will learn.
But the answer to the question if $\mathcal{T}_1 \subsetneq \mathcal{T}_2$ are two strictly ordered topologies on a set $X$, must there always be a sequence in $X$ with $x_n \to x$ in $\mathcal{T}_1$ but not $x_n \to x$ in $\mathcal{T}_2$ is not necessarily true. It will only hold if both topologies are sequential. (Say metric or first countable, like the reals and the lower limit topology).
A counterexample is the reals in $\mathcal{T}_{cc}$ the co-countable topology and the strictly larger $\mathcal{T}_d$ the discrete topology. These topologies have the exact same convergent sequences ($x_n \to x$ iff $\exists N: \forall n \ge N: x_n =x$) but one is really larger than the other. So a finer topology need not make fewer sequences converge, in general.
A: Your claim holds if and only if the topology is a sequential space. That is to say, a topology $\tau$ is sequential if and only if all finer topologies have fewer sequential limits. Stated differently, a topology $\tau$ admits a strictly finer topology with exactly the same sequential limits if and only if $\tau$ is not sequential.
This relies on a particular fact: if $\tau$ is a topology, and $\pi$ is the set of all sequentially open sets of $\tau$, then $\pi$ is a topology with $\tau\subseteq \pi$, and a sequential limit holds in $\pi$ if and only if it holds in $\tau$. From this it follows logically that $\pi$ is exactly the set of its own sequentially open sets, and is therefore a sequential space.
Using the above fact, take any topology $\tau$. If $\tau$ is not sequential, then necessarily $\tau\neq\pi$ since $\pi$ is sequential, thus $\pi$ is a strictly finer topology with exactly the same sequential limits. Conversely if $\tau$ is sequential and $\tau'$ is strictly finer than $\tau$, then $\pi'$ is sequential and strictly finer than $\tau$. Since $\tau$ and $\pi'$ are distinct sequential spaces, they cannot have exactly the same sequential limits, so likewise $\tau$ does not have exactly the same sequential limits as $\tau'$, so no strictly finer topology can have the same sequential limits as $\tau$.
