Suppose that $S$ is an embedded/regular submanifold of $M$ with $\mathrm{dim}\ S = s < \mathrm{dim}\ M$. If $U$ is an open subset of M, then $S' = U \cap S$ is an open subset of $S$ in the subspace topology.
Question: If $S' \neq \varnothing$, is the dimension of $S'$ known?
If $S' \neq \emptyset$ then it is a nonempty open subset of $S$ and therefore has dimension $s$. This is completely general - if $M$ is a manifold and $\emptyset \neq U \subseteq M$ then $\dim U = \dim M$. For instance, an atlas on $M$ witnesses local diffeomorphisms of $M$ with open subsets of $\mathbb R^n$ where $n = \dim M$. Restricting this atlas to $U$ yields local diffeomorphisms of $U$ with open subsets of $\mathbb R^n$, so $\dim U = n = \dim M$. You could also look at tangent spaces to conclude the same thing.