To show that $\mathscr{B}$ is a base for a topology on $X$, you must show two things:
- For each $x\in X$ there is some $B\in\mathscr{B}$ such that $x\in B$, and
- for any $B_0,B_1\in\mathscr{B}$ and any $x\in B_0\cap B_1$, there is some $B\in\mathscr{B}$ such that $x\in B\subseteq B_0\cap B_1$.
In your case $\mathscr{B}=\mathscr{B}_1\cup\mathscr{B}_2$, where $\mathscr{B}_1=\Big\{\{n\}:n\in X\text{ and }n>0\Big\}\;,$ and $\mathscr{B}_2=\{X\setminus F:F=\{1,\dots,n\}\text{ for some }n\in\Bbb N\}\;.$
If $n\in X\setminus\{0\}$, then $n\in\{n\}\in\mathscr{B}_1\subseteq\mathscr{B}$, and $0\in X\setminus\{1\}\in\mathscr{B}_2\subseteq\mathscr{B}$, so condition (1) is satisfied.
Condition (2) is also easy to check. Suppose that $B_0,B_1\in\mathscr{B}$ and $n\in B_0\cap B_1$. If $B_0$ and $B_1$ both belong to $\mathscr{B}_1$, then clearly $B_0=B_1=\{n\}$, and in (2) we can take $B$ to be $B_1$. In fact, you can easily check that if either of $B_0$ and $B_1$ belongs to $\mathscr{B}_1$, we can take that set to be the $B$ of (2): if a member of $\mathscr{B}_1$ intersects a member of $\mathscr{B}$, it must already be a subset of that member. Thus, the only case that requires a little work is when $B_0,B_1\in\mathscr{B}_2$, and I leave you to check that in that case one of them is again a subset of the other.
To show that the resulting topology is Hausdorff ($T_2$), you need only show that if $m$ and $n$ are distinct elements of $X$, there are $B_m,B_n\in\mathscr{B}$ such that $m\in B_m$, $n\in B_n$, and $B_m\cap B_n=\varnothing$. This is dead easy of neither $m$ nor $n$ is $0$: just take $B_m=\{m\}$ and $B_n=\{n\}$. Separating $0$ and some $n>0$ is almost as easy: for you neighborhood of $n$ you will of course choose $\{n\}\in\mathscr{B}_1$, and for your neighborhood of $0$ you'll want some member of $\mathscr{B}_2$ that misses $n$. The simplest choice is $X\setminus\{1,\dots,n\}$, but any $X\setminus\{1,\dots,m\}$ with $m\ge n$ will work just as well.
You should try to convince yourself that this topology on $X$ makes $\langle 1,2,3,\dots\rangle$ a sequence converging to the point $0$. In fact, as Martin noted in the comments, $X$ looks just like the subset of $\Bbb R$ consisting of $0$ and the reciprocals $1/n$ for 0. Call that subset $Y$. If you take as a base the sets $\{1/n\}$ for integers $n>0$ and the sets $Y\setminus\{1,\frac12,\dots,\frac1n\}$ for integers $n>0$, imitating $\mathscr{B}_1$ and $\mathscr{B}_2$ in your problem, you get the same topology on $Y$ that it inherits from the usual topology on $\Bbb R$, in which $\langle 1,\frac12,\frac13,\dots\rangle$ converges to $0$.