This is Exercise 5.20 from John Lee's ISM. The text says this is simply an observation but I'm having trouble proving this fact.
Suppose $M$ is a smooth manifold and $S \subset M$ is an immersed submanifold. Show that every subset of $S$ that is open in the subspace topology is also open in its given submanifold topology; and the converse is true if and only if $S$ is embedded.
$S\subset M$ immersed submanifold means that $S$ is endowed with a topology (call it the submanifold topology) and a smooth structure in which the inclusion map $S \hookrightarrow M$ is a smooth immersion.
Since the subspace topology is the coarsest topology in which the inclusion map is continuous, and smooth maps are continuous the first fact follows. However, I am not sure how to show that the submanifold topology is contained in the subspace topology only if $S$ is embedded.
$S$ is an embedded submanifold if and only if the inclusion $S\hookrightarrow M$ is a smooth embedding, but we are supposing already that $S\hookrightarrow M$ is a smooth immersion, so it follows that $S$ is an embedded submanifold (in our given scenario) if and only if $S\hookrightarrow M$ is a topological embedding.
Okay now, if you unravel the definition a tiny bit you'll see that $S\hookrightarrow M$ is a topological embedding if and only if the given submanifold topology on $S$ is the same as the subspace topology coming from $M$. But we are in a situation where you've already proved that the submanifold topology is finer than the subspace topology, so in fact the two topologies agree if and only if the subspace topology is finer than the submanifold topology