Longest sequence of minimally finer topologies Suppose we start with a topology $T_1$ of X. Is there a way to get construct a sequence of topologies $T_n$ such that $T_{n - 1} \subset T_{n}$ in which there is no finer topologies in between, also that sequence is the longest one‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌.
 A: Suppose the topology $T_1$ is Hausdorff and sequential.  Then I claim that such a longest sequence exists rather vacuously: namely, there does not even exist a topology $T_2$ such that $T_1\subset T_2$ and there is no topology in between.  That is, a Hausdorff sequential topology admits no minimal refinement.
To prove this, suppose such a $T_2$ did exist.  Since $T_1$ is sequential, it is the finest topology for which all of the $T_1$-convergent sequences converge.  Since $T_2$ is finer than $T_1$, this means there is some sequence $(x_n)$ which converges to some $x\in X$ with respect to $T_1$ but not with respect to $T_2$.  Thus some $T_2$-neighborhood of $x$ excludes infinitely many of the $x_n$; passing to a subsequence, we may assume that $(x_n)$ does not even accumulate at $x$.  Since $T_1$ is Hausdorff, $(x_n)$ cannot accumulate anywhere else, so this means that the set $C=\{x_n\}$ is $T_2$-closed.  By minimality of $T_2$, this means that $T_2$ is the topology generated by $T_1$ together with $X\setminus C$.  But if $D\subset C$ is any infinite coinfinite set, then $D$ is also $T_2$-closed, and the topology $T$ generated by $T_1$ and $X\setminus D$ is satisfies $T_1\subset T\subset T_2$ (it is easy to see that any $T$-neighborhood of $x$ must contain all but finitely many points of $C\setminus D$).  This contradicts the minimality of $T_2$.
In brief: any minimal refinement of a sequential Hausdorff topology must be obtained by taking a convergent sequence and making it not converge, but then you can get a smaller refinement by choosing a proper subsequence to make not converge.
Both the Hausdorff and sequential hypotheses are necessary here.  For a counterexample which is sequential but not Hausdorff, let $T_1$ be the cofinite topology on any infinite set (you can then take $T_2=T_1\cup\{\{x\}\}$ for any $x\in X$; this example also shows that the Hausdorff hypothesis cannot be weakened to the hypothesis that points are closed).  For a counterexample where $T_1$ is Hausdorff but not sequential, let $F$ be a nonprincipal ultrafilter on $\mathbb{N}$, $X=\mathbb{N}\cup\{\infty\}$, and $T_1$ consist of all subsets of $\mathbb{N}$ together with all sets of the form $\{\infty\}\cup A$ for $A\in F$.  From the fact that $F$ is an ultrafilter, it is easy to see that the discrete topology $T_2$ is a minimal refinement of $T_1$.  (This then gives another example where a longest sequence of minimal refinements of $T_1$ exists, where the length of the sequence is now $2$.  A similar construction using $n$ different ultrafilters gives a Hausdorff topology such that there is a longest sequence of minimal refinements of length $n+1$, for any $n\in\mathbb{N}$.)
A: There's no longest such sequence, though for a given set $X$ there are (using the Axiom of Choice) maximal such chains. You can construct a chain of topologies with the successor property you state that is at least as long as the cardinality of $X$.
Revision
@bof pointed out a flaw in my original construction (presented below for context), in place of which I offer this even simpler one:
Let $(x_{\iota})_{{\iota < \kappa}}$ enumerate $X$, where $\kappa = \lvert X \rvert$. For $\alpha \le \kappa$, let $X_{\alpha} = \{x_{\iota} \mid\ \iota < \alpha\}$, so $X_0 = \emptyset$; for $\beta \le \kappa$, let $T'_{\beta} = \{X_{\alpha} \mid\ \alpha \le \beta\} \cup \{X\}$. It's easy to see that each $T'_{\beta}$ is a topology on X – something like a "partial initial segment" topology (with $X$ wellordered by the enumeration, $X_{\alpha}$ is just $[x_0, x_{\alpha})$ ). Furthermore, $T'_{\beta +1} = T'_{\beta} \cup \{X_{\beta + 1}\}$, so clearly $T'_{\beta +1}$ is a minimal refinement of $T'_{\beta}$. At the last stage, $T'_{\kappa}$ is the (full) initial segment topology, consisting of all and only all the $X_{\alpha}, \alpha < \kappa$. None of these topologies is Hausdorff.
The length of the sequence is $\kappa + 1$, but it's possible to keep going. For example, adding $\{x_1\}$ to $T'_{\kappa}$ gives a topology that's a further minimal refinement.
The order topology is a refinement of $T'_{\kappa}$ though certainly not minimal, and I'm not sure (yet?) if it can be reached by a chain of minimal refinements.
Original, flawed construction
Define a chain of topologies $T_{\alpha}, \alpha \le \kappa$ as follows: $T_0 = \{\emptyset, X\}$ is the indiscrete topology; $T_{\alpha +1}$ is the topology generated by $T_{\alpha}$ together with the singleton $\{x_{\alpha}\}$; at limit stages $\lambda \le \kappa$, let $T_{\lambda}$ be the topology generated by the union $\bigcup_{\alpha < \lambda} T_{\alpha}$.
It's not hard to see that if $\alpha < \beta$ then $T_{\alpha} \subsetneq T_{\beta}$, and that the chain ends with $T_{\kappa}$ = the discrete topology. The length of the chain is the ordinal $\kappa + 1$.
I had in mind infinite sets $X$, but as with the revised construction, there's no such restriction. If $X = \{0, 1\}$, then $T_0$ is indiscrete, $T_1$ is Sierpinski space, and $T_2$ is discrete.
The flaw which @bof noted is that $T_{\alpha +1}$ is not a minimal refinement of $T_{\alpha}$. For example, if $|X| > 2$, then in between $T_1 = \{\emptyset, \{x_0\}, X\}$ and $T_2 = \{\emptyset, \{x_0\}, \{x_1\}, \{x_0, x_1\}, X\}$ lies a topology $T_{1.5} = \{\emptyset, \{x_0\}, \{x_0, x_1\}, X\}$. Note that $T_{1.5} = T'_2$ of the revised construction.
