Countability, separability and Lindelöf of $[-1,1]$ with another topology.

I am trying to solve the following exercise.

Study the countability axioms, separability axioms and the Lindelöf property in $(X,\tau)$, where $X=[-1,1]$ and $U\in\tau$ if and only if $0\notin U$ or $(-1,1)\subset U$.

Is this the excluded point topology in $[-1,1]$, excluding $0$? I think it is, since the only open set including $0$ is $X$. But I don't know if this is the right conclusion.

If it is, I know that the excluded point topology es second-countable, $T_0$ and Lindelöf (because it is second-countable).

Is this right or I am leaving something?

A subset is open when we know $0 \notin U$ or $(-1,1) \subseteq U$. So it's not the excluded point topology but it's almost that: the only extra open sets beyond the excluded point topology wrt $0$ are $(-1,1), (-1,1], [-1,1)$.
So already all subsets of $[-1,1]\setminus \{0\}$ are open, so this forms a discrete subspace of size continuum, so $X$ is certainly not second countable: any base must at least contain all sets $\{x\}$ for $-1 \le x < 0$ and $0 < x \le 1$. As any dense subset of $X$ must also intersect all these singletons, $X$ is not separable.
If we have an open cover of $X$, to cover $0$ we need one of the above extra open sets (or $X$) and then we only need to cover $-1$ or $1$ or both to have a finite subcover. So $X$ is compact so certainly Lindelöf. We cannot separate $0$ from $\frac12$ simultaneously, or note that $\{\frac12\}$ is not closed, to see that $X$ is neither $T_1$ nor $T_2$. I do think $X$ is normal (some boring case distinctions).