Exercise from Terence Tao In the exercise below I don't understand why Exercise 13.5.9 doesn't contradict 13.5.8 ?
 A: Exercise 13.5.9 is wrong as stated.  A topological space in which every sequence has a convergent subsequence is said to be sequentially compact.  Not every compact space is sequentially compact.
EDIT: Exercise 13.5.8 is also wrong.  This is not compact!  Let $S = \{x_i: i \in \mathbb N\}$ be a sequence of distinct points of $X \backslash \{\infty\}$, and consider the open cover $\{U_i: i \in \mathbb N\}$ where $U_i = S^c \cup \{i\}$.  This has no finite subcover.
A: There are corrections here to both of these exercises.

p. 390: Exercise 13.5.8 should be replaced as follows: “Show that
  there exists an uncountable well-ordered set $\omega_1+1$ that has a
  maximal element $\infty$, and such that the initial segments $\{ x \in\omega_1+1: x < y \}$ are countable for all $y \in \omega_1+1 \backslash \{\infty\}$.   (Hint: Well-order the real numbers using Exercise
  8.5.19, take the union of all the countable initial segments, and then adjoin a maximal element $\infty$.)  If we give $\omega_1+1$ the order
  topology (Exercise 13.5.5), show that $\omega_1+1$ is compact; however,
  show that not every sequence has a convergent subsequence.”

and

p. 438, Exercise 13.5.9: One needs to assume as an additional
  hypothesis that X is first countable, which means that for every x in
  X there exists a countable sequence V_n of neighborhoods of x, such
  that every neighbourhood of x contains one of the V_n.

