$U\subset X$ is open if and only if for each $x \in U$ and every sequence $(x_n)_{n=1}^\infty$ of points in $X$ converging to $x \in U$, there exists $N>0$ such that for all $n \ge N$ implies $x_n \in U$.
How/Why does this condition define a topology? Does this topology have a name?
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As I pointed out in this recent answer, this is one of the standard ways to define a topology if we have just a set (without a topology yet) with a "notion of convergence" (a so-called convergence space).
If we are given a topology, your statement is the definition of convergence of sequences.
If we start with a convergence notion, define the topology and see what sequences converge for that new topology we might or might not get more convergence sequences. If we get exactly the original ones back then we have a"topolgoical convergence space" and the topology that results is called a "sequential space".
A classic example of a non-topological convergence notion in the above sense is "convergence almost everywhere" for functions wrt the Lebesgue measure $\mathbb{R}$, though this is rather hard to show here now.
As pointed out in the comments, this doesn't define a topology on its own. What could it mean to talk about sequence convergence without a notion of some topology? There isn't a natural way to do it.
To demonstrate why, let's consider the real line $X=\Bbb R$ as the underlying set, and consider the following four topologies on $X$:
Now, consider the sequences $x_n:=\frac1{n+1}$ and $y_n:=1.$ It seems like the first sequence should converge to $0$ (and nowhere else), while the second should converge to $1$ (and nowhere else). However, we're in for a surprise.
In the first and fourth topologies, the first sequence doesn't converge at all (though the second one converges as we expected). In fact, the only sequences that can converge in those cases are the ones that are eventually constant.
In the second topology, the first sequence converges to every point of $x$; the second sequence fails to converge! Only the sequences that would converge to $0$ in the usual topology can converge in the second topology, and each of these converges to all points of $X.$
In the third topology, the first sequence fails to converge, and the second sequence converges to all points of $X,$ reasoning similarly to the second topology.
If a topological space satisfies the condition you post, it is known as a sequential space.
Not all topological spaces are sequential. The space $X$ under the fourth topology above is an example of a non-sequential space. To see why, consider the set $U=(-1,1).$ It is sequentially open (since any sequence that converges to a point in $U$ is eventually constant, meaning we can find such an $N$), but not open, as it is non-empty and has an uncountable complement.