Let $M_1,M_2 \subset \Bbb R^3$ smooth and compact manifolds such that $M_1$ has dimension $1$ and $M_2$ has dimension $2$. Assume $M_1 \cap M_2$ is an infinte set. Prove that there exists $x\in M_1 \cap M_2$ such that $$T_xM_1 \subset T_xM_2$$
I'm having trouble approaching to this. My intuition tells me that there exists a neighborhood $U$ such that $M_1\cap U \subset M_2 \cap U$, and then the claim follows. However I'm not sure if it's correct/how to prove it. Any ideas?
First take any $x\in M_1\cap M_2$. If $T_xM_1\not\subset T_xM_2$, then, because $T_xM_1$ is one-dimensional, this implies that $T_xM_1\cap T_xM_2=\{0\}$ and hence $T_x\mathbb{R}^3=T_xM_1 \oplus T_xM_2$. In other words, $M_1$ and $M_2$ are transverse at $x$, which implies that $x$ is an isolated point of $M_1\cap M_2$, i.e. there is a neighbourhood $U$ of $x$ in $\mathbb{R}^3$ such that $U\cap M_1\cap M_2=\{x\}$.
Now, we can show that if $M_1$ and $M_2$ are compact and $M_1\cap M_2$ is infinite, then there is $x\in M_1\cap M_2$ such that $T_xM_1\subset T_xM_2$. Take any sequence $x_n\in M_1\cap M_2$ of distinct points. By compactness of $M_1\cap M_2$, this sequence has a subsequence which converges to some $x\in M_1\cap M_2$. If $T_xM_1\not\subset T_xM_2$, then the first paragraph shows that $x$ is an isolated point of $M_1\cap M_2$, which contradicts that $x$ is the limit of a sequence of distinct points in $M_1\cap M_2$. Hence, $T_xM_1\subset T_xM_2$.