A wrong counterexample for the Sequential Compactness Theorem for general metric space? Example 12.38 from the book Advanced Calculus by Patrick M. Fitzpatrick is trying to say that the assertion "sequentially compact if and only if X is a closed bounded" does not necessarily hold for all metric spaces. 
The counterexample is the subset of $C([0, 1], \mathbb{R})$ consisting of all continuous  functions such that $|f(x)| \le 1$ for all $x$ in $[0,1]$. It then claims that this subset is closed and gives the sequence of functions $f_k(x)=x^k$ as a counterexample for the subset not to sequentially compact. 
But the problem is that this subset is not closed because the same sequence of functions $f_k(x)=x^k$ converges to the function that satisfy $|f(x)| \le 1$ for all $x$ in $[0,1]$ but it is not in $C([0, 1], \mathbb{R})$ contradiction to the definition of closedness of a set. Am I right?
 A: Not quite. When you say "the sequence of $x^k$ converges to something" what topology do you mean? You certainly do not mean uniform convergence (in sup-norm), because the sequence $x^k$ is not Cauchy sequence in sup-norm. Do you mean the point-wise (weak) convergence? When the author says that the unit ball is closed, he means the original (normed/strong) topology. In this topology, the unit ball is closed because any sequence of functions that converges uniformly will have the limit to be a continuous function with the norm being bounded by one (norm is continuous in the original topology). A sequence cannot converge to something outside the space if the space is complete ($C[0,1]$ with the sup-norm is).
A: No, you're not right. In the metric $d(f,g) = \sup \{|f(x) - g(x)|: x \in [0,1]\}$ that is used on this space $X= C([0,1],\mathbb{R})$, the sequence $f_k$ does not converge at all. So it cannot be a counterexample to closedness of the set $C= \{f: \forall x: |f(x)| \le 1\}$, because then there has to be a function $f \in X$ (!) such that $f_k \to f$ with $f_k \in C$ for all $k$ and $f \notin C$, because we are talking about closed sets in $(X,d)$.
Suppose $f_k$ did converge to some $f \in X$. Then it has to converge pointwise to $f$ too, as $|f_k(x) - f(x)| \le d(f_k,f) \to 0$ for any fixed $x \in [0,1]$, and this forces $f(0) = 0$ for $x \in [0,1)$ and $f(1) = 1$, which contradicts the continuity of $f$. 
$C$ is bounded by definition, as it is just the closed ball of radius $1$ around
the $0$-function, it is closed too: suppose $g \notin C$, then there is some $p \in [0,1]$ with $|f(p)| > 1$. Then it's easy to check that, setting $r = 1-|f(p)| >0$, $B_d(g,r) \cap C =\emptyset$. 
The fact that no subsequence of $(f_k)$ converges (the argument I gave also works for any subsequence) shows that $C$ is not (sequentially ) compact, as claimed.
