Is the topologist's sine curve a manifold? [duplicate]

The topologist's sine curve $S$ is a subspace of $\mathbb{R}^2$ meaning that it is a subset of $\mathbb{R}^2$ and inherits its topology from the topology of $\mathbb{R}^2$. In order to be a manifold it must be locally Euclidean in the inherited topology. That means that it must also be locally connected but, as noted, it is not locally connected on $0\times[-1,1]$.

You ask whether perhaps some other topology on $S$ would make it a manifold.

Perhaps, but with some topology other than the one it inherits from $\mathbb{R}^2$ it is no longer the topologist's sine curve.

Choosing a specific subset $I$ and showing that there's no coordinate chart for $I$ is not enough. However, you are indeed very close with your idea.

If $S$ is a manifold, then by definition the point $p=(0,0)$ must have a neighborhood which is homeomorphic to $\Bbb{R}^n$. However, $\Bbb{R}^n$ is path connected, and all neighborhoods of $p$ are not. Path connectedness is preserved by homeomorphism. Therefore $S$ cannot be a manifold.

You can consider the same set with a different topology, but as John has pointed out, you are no longer talking about the Topologist's Sine Curve anymore. Any intuition you have about what that topological space "looks like" goes out the window if you introduce a different topology.

For instance, you can give $S$ the discrete topology, which means every subset of $S$ is open. Then for every $p\in S$, $\{p\}$ is a neighborhood of $S$ which is homeomorphic to $\Bbb{R}^0$, so it is a $0$-manifold.