In the curves and surfaces (Second edition) book from Montiel-Ross, we have the next proposition and proof:
Proposition: Let $R^{+}$ be a straight line whose origin is at $\mathbb{R}^3-S$, which intersects a compact surface $S$ transversally. If $\{R_{n}^{+}\}_{n\in \mathbb{N}}$ is a sequence of lines converging to $R^{+}$ starting at the points of $\mathbb{ R}^3-S$, then there exists a number $N\in\mathbb{N}$ such that, if $n\geq N$, $R_{n}^{+}$ cuts $S$ transversely and the number of points in the sets $R_{n}^{+}\cap S$ and $R^{+}\cap S$ coincide.
Proof:
The transversality of the intersection and the compactness of $S$ guarantee us that $S\cap R^{+}$ is a finite set. Let $$S\cap R^{+}=\{p_{1},\dots, p_{k}\}.$$ Suppose the statement about the sequence $\{R_{n}^{+}\}_{n\in \mathbb{N}}$ is false. In that case, we can assume, using a subsequence of the original, that the sequence $\{R_{n}^{+}\}_{n\in \mathbb{N}}$ of lines starting at the points of $\mathbb{R}^3-S$ and convergent to $R^{+}$ is formed by lines that satisfy at least one of the following conditions:
A.- $R_{n}^{+}$ is tangent to $S$ at some point.
B.- The number of points of $R_{n}^{+}\cap S$ is greater than the number of points of $S\cap R^{+}$.
C.- The number of points of $R_{n}^{+}\cap S$ is less than the number of points of $S\cap R^{+}$.
Now, the book proves that any of the above conditions leads to a contradiction. In the A case, it says that there exists a secuence of points $p_{n}\in S\cap R_{n}$ such that $R_{n}$ would be tangent to $S$ at each $p_{n}$, $n\in \mathbb{N}$. Since $S$ is compact, it says that this sequence converges to a point $p\in S$, which will also belong to $R$. Why is it that we have full convergence of this sequence, instead of the convergence of a subsequence of the original one?