(5.27) What does convergence mean for a sequence in a topological space? It means that for any open set $U$ containing $x$, the points $x_n$ are in $U$ for all sufficiently large $n$ (this recovers the usual metric notion of convergence if you take $U$ to be a sequence of balls of smaller and smaller radius going to zero).

(6.40) Suppose that we have a sequence $x_n$ such that $x_n\to x$ and let $y$ be a point with $x\neq y$. By the Hausdorff assumption, there are disjoint open sets $U\ni x$ and $V\ni y$. Because $x_n\to x$ there exists an $N$ such that $x_n\in U$ for all $n\geq N$. Therefore $x_n\not\in V$ for all $n\geq N$, and hence $x_n$ does not converge to $y$.

We need to show that $X\setminus K$ is open. If $X\setminus K=\emptyset$ then it's open, so assume that it is nonempty. Pick a point $y\in X\setminus K$. For each $x\in K$ there is a ball $U_x\ni x$ and $V_x\ni y$ such that $U_x\cap V_x=\emptyset$ (by the Hausdorff assumption). We would like to say $\bigcap_{x\in K} V_x$ is an open neighbourhood of $Y$ disjoint from $K$. Certainly it is disjoint from $K$ (otherwise there is some point $x\in V_x$, but $V_x\cap U_x=\emptyset$ and $x\in U_x$) but it might not be open because it could be an infinite intersection.

(11.11) As $K$ is compact, there is a finite collection $\{x_i\ :\ i\in I\}$ (for a finite set $I$) such that $\{U_{x_i}\ :\ i\in I\}$ covers $K$. Now the intersection $V=\bigcap_{i\in I}V_{x_i}$ is open (as it's a finite intersection). Moreover $V\cap K=\emptyset$.

(13.30) Therefore $K$ is closed (because its complement is a union of open sets $V$ like we just constructed, hence open).