Just a basic pointset topology question: clearly we can detect differences in topologies using convergent sequences, but is there an example of two distinct topologies on the same set which have the same convergent sequences?
In a metric (or metrizable) space, the topology is entirely determined by convergence of sequences. This does not hold in an arbitrary topological space, and Mariano has given the canonical counterexample. This is the beginning of more penetrating theories of convergence given by nets and/or filters. For information on this, see e.g. http://math.uga.edu/~pete/convergence.pdf In particular, Section 2 is devoted to the topic of sequences in topological spaces and gives some information on when sequences are "topologically sufficient". In particular a topology is determined by specifying which nets converge to which points. This came up as a previous MO question. It is not covered in the notes above, but is well treated in Kelley's General Topology. 


The cocountable topology on an uncountable set is undistinguishable from the discrete topology if you can only use sequences. 


Just another example. Consider the Banach space $\ell^{1}\left(\Gamma\right)$ , $\Gamma$ being an infinite set. Then the weak topology and the norm topology have the same convergent sequences (Schur' Theorem), while they are clearly distinct. 


There is a category of "sequential spaces" in which objects are spaces defined by their convergent sequences and morphisms are those maps which send convergent sequences to convergent sequences. As stated above, all metric spaces are sequential spaces, but so are all manifolds, all finite topological spaces, and all CWcomplexes. To build this category, one actually just needs to look at the category of right $M$sets for a certain monoid $M$. Consider first the "convergent sequence space" $S:=${$\frac{1}{n}n\in{\mathbb N}\cup${$\infty$}}$\subset {\mathbb R}$. In other words $S$ is a countable set of points converging to 0, and including $0$. Let $M$ be the monoid of continuous maps $S\to S$ with composition. Then an $M$set is a "set of convergent sequences" closed under taking subsequences. The category of $M$sets is a topos, so it has limits, colimits, function spaces, etc. And every $M$set has a topological realization which is a sequential space. 


I like the following example. Let $D$ be an infinite discrete space and $\beta D$ its StoneČech compactification. Then a sequence in $\beta D$ converges if and only if it is eventually constant. Thus convergent sequences do not distinguish between the compact topology of $\beta D$ and the discrete topology on its underlying set. Obviously  the same holds for every refinement of $\beta D$. 


In the space $[0,\omega_1]$ with the order topology the point $\omega_1\in\overline{[0,\omega_1)}$, but for every convergent sequence $(x_n)_{n\in\Bbb N}\subset[0,\omega_1)$, $$\lim_{n\to\infty}x_n<\omega_1.$$ 

