This question came from more specific question, that I am trying to solve with the above statement.
The specific question is : Given $M_1, M_2$ in $\mathbb{R}^3$ , smooth compact manifolds, while $M_1$ is 1-dimensional and $M_2$ is 2-dimensional. Assuming that $M_1 \cap M_2$ is infinity prove that there exists $x \in M_1 \cap M_2$ such that $TxM_1 \subset TxM_2$.
I managed to prove existence of x, such for any neighborhood of x, $Ux$.
$M_1 \cap M_2 \cap Ux \neq \{x\}$ , aka for any neighborhood of x, there is point $y$, such that $ x \neq y \in M_1 \cap M_2 \cap Ux$. could not get any further.
If $x \in M_1 \cap M_2$ is such that $T_xM_1 \oplus T_x M_2 = T_x \Bbb R^3$ then there is a neighborhood $U$ of $x$ such that $M_1 \cap U \cap M_2 = \{x\}$. Since $M_1$ is compact it follows that there is only finitely many such $x$.
(Notice that compactness of $M_2$ is not needed.)
Why there is such a neighborhood ? There a neighborhood $V \ni x$ with local coordinates $a,b,c$ around $x$ such that $M_2 \cap V = \{(a,b,c) \mid c = 0 \}$. Consider a parametrization of $M_1$ given by $\gamma(t) = (a(t),b(t), c(t))$, with $\gamma(0) = x$. On has $c(0) = 0$ by hypothesis, and since $T_xM_1 + T_xM_2 = T_x \Bbb R^3$ we should have $c'(0) \neq 0$. In particular, for $ t> 0$ small enough, $c(t) \neq 0$ so $\gamma(t) \cap M_2 = \emptyset$ and it follows that there is a neighborhood $U \ni x$ with $M_1 \cap M_2 \cap U = \{x\}$.