Problem: Suppose $S$ is a subset of a manifold $M$, and for every point $p \in S$, there exists an open neighbourhood $U$ around $p$ such that $S \cap U$ is a submanifold of the neighbourhood $U$. Show that $S$ is in fact a submanifold of the manifold $M$. This problem is quite similar to Lemma 5.5 in Lee's smooth manifold book (which he did not prove).
Attempt: Pick $p \in S$ and find the corresponding neighbourhood $U$. We can easily extend the submanifold charts $\{U_{\alpha} \cap U,\phi_{\alpha}|_U \}$ for $U$ (looking at $U$ as a submanifold of $M$) to the whole manifold $M$, and if we are lucky enough to have the $U_{\alpha} \cap S \subset U \cap S$ then we are good for the point $p$. But if $U \cap S \subsetneq U_{\alpha} \cap S$, I cannot say that $\phi_{\alpha}(U_{\alpha} \cap S) = \phi_{\alpha}(U_{\alpha}) \cap \mathbb{R}^k \subset \mathbb{R}^n$, so I need to find some way to "glue" the remaining parts of $U_{\alpha} \cap S$ onto the set while proving that $\phi_{\alpha}(U_{\alpha} \cap S) = \mathbb{R}^k \cap \phi_{\alpha} (U_{\alpha})$. I am stuck at this step, because simply finding another neighbourhood that actually covers $(U_{\alpha} \cap S) - (U \cap S)$ and coming up with charts on it just give me diffeomorphisms by composing the charts, and I am not sure whether that could help me proving that $\phi_{\alpha}$ does "straighten out" the $(U_{\alpha} \cap S) - (U \cap S)$. Any help is appreciated!
You don't need to have $U_\alpha\cap S\subset U\cap S$, if $U_\alpha$ is a chart, every connected component of $U_\alpha\cap U$ is a chart. Take the connected component $V_\alpha$ which contains $p$ and restrict $\phi_\alpha$ to $V_\alpha$.