Let me give first the definitions:
D A metric space $M$ (or subset of a metric space) is said to be precompact if every sequence in $M$ contains a Cauchy (or fundamental) subsequence.
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D Let $S$ be any subset of $\Bbb R^n$. Given $\epsilon >0$, we say that $N$ is an $\epsilon$-net for $S$ if the set of open balls
$$B_\epsilon(N)=\{B(x,\epsilon):x\in N\}$$
covers $S$. That is, the set of open balls of radius $\epsilon$ centered at the points of $N$ cover $S$.
Now, a theorem by Hausdorff
T Let $P$ be a metric space, $M$ a subset of $P$. Then $P$ is precompact if and only if, given $\epsilon >0$, $P$ contains a finite $\epsilon$-net for $M$.
P Suppose $M$ is precompact. Then given $\epsilon >0$, choose $x_1\in M$. If the point $x_1\in M$ is such that $d(x,x_1)<\epsilon$ for each $x\in M$; $\{x_1\}$ is itself an $\epsilon$-net, and we're done. If there exists points with $d(x,x_1)>\epsilon$, choose as $x_2$ one of these points. If now for every $x\in M$ we have either $d(x,x_1)<\epsilon$ or $d(x,x_1)<\epsilon$ , we're done. Else, continue the process, noting that the distance from each new $x_n$ to the $x_i\; ;1\leq i\leq n-1$ exceeds $\epsilon$. Thus, if the construction fails to terminate after a finite number of steps, we would obtain a sequence $\{x_n\}_{n\in \Bbb N }$ which contains no Cauchy subsequence, contrary to the hypothesis that $M$ is precompact. Thus, the construction must terminate, providing us with a finite $\epsilon$-net, $\{x_1,\dots,x_{\ell}\}$.
Conversely, suppose that for any $\epsilon>0$, $P$ contains a finite $\epsilon$-net for $M$, and let $A$ be any infinite subset of $M$, in particular, a sequence containing infinitely many distinct points of $M$. We can select a Cauchy subsequence as follows. Let $x_0\in A$ be any point, then take $\epsilon =1$ in the condition and cover $A$ with a finite number of balls of radius $1$. One of these balls, say $B_1$, must contain an infinite $A_1$ of $A$. Choose $x_1\in A_1\setminus \{x_0\}$. Choosing $\epsilon = 1/2$; we cover $A_1$ with a finite number of balls of radius $1/2$. One of them, say $B_2$ must contain an infinite subset $A_2$ of $A_1$. Choose $x_2\in A_2\setminus \{x_0,x_1\}$. Continuing this process, we obtain a sequence $A_0\supseteq A_1\supseteq A_2\cdots$, where each $A_n$ is contained in a ball of radius $1/n$ and a sequence of distinct points $x_0,x_1,x_2,\dots$ with $x_n\in A_n$. This sequence is Cauchy, for if $m<n$, $$B_m\supseteq A_m\supseteq A_n$$ which means $$d(x_n,x_m)<\frac 2 m$$Since $2/m\to 0$, we're done. It follows $M$ is precompact.
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Look here for
T Let $S$ be bounded in $\Bbb R^n$. Then $S$ is precompact.
Now, since $\Bbb R^n$ is complete, any Cauchy (sub)sequence will converge. The assumption that the set is closed means the limit $\ell$ will be in such set. Thus, every bounded closed subset of $\Bbb R^n$ contains a convergent subsequence, that is, it is sequentially compact.
NOTE Observe the similarity of the $\implies$ direction of Hausdorff's criterion with the proof of Weiertrass-Bolzano that Ragib suggests.
Compactness does not imply sequential compactness.
Compactness implies that every sequence has an accumulation point, which is equivalent to countable compactness [every countable open cover has a finite subcover]. But in general, a sequence having accumulation points does not imply that the sequence has a convergent subsequence. One needs additional hypotheses, e.g. first countability of the space to have that implication.
One example of a space that is compact but not sequentially compact is, as shown by the example, the closed unit ball of $(\ell^\infty)^\ast$ in the weak$^\ast$ topology.
A perhaps easier to visualize example is a product of sufficiently many copies of $\{0,1\}$. (Any example must be somewhat difficult to visualize, since the easy-to-visualize spaces have a strong tendency to be first-countable.)
Let $\mathscr{P}(\mathbb{N})$ denote the power set of $\mathbb{N}$, and $X = \{0,1\}^{\mathscr{P}(\mathbb{N})}$ (that is up to the naming of the indices $\{0,1\}^{\mathbb{R}}$, but taking $\mathscr{P}(\mathbb{N})$ makes it easier to define a sequence without convergent subsequences). Define the sequence $(x_n)_{n\in\mathbb{N}}$ in $X$ by
$$p_M(x_n) = \begin{cases} 0 &, n \notin M\\
0 &, n\in M \text{ and } \operatorname{card} \{m\in M : m < n\} \text{ even}\\
1 &, n\in M \text{ and } \operatorname{card} \{ m\in M : m < n\} \text{ odd},\end{cases}$$
where $p_M \colon X \to \{0,1\}$ is the coordinate projection. Then $(x_n)_{n\in\mathbb{N}}$ has no convergent subsequences. For if $(x_{n_k})_{k\in\mathbb{N}}$ is a subsequence, consider the set $M = \{ n_k : k\in\mathbb{N}\}$. Then $p_M(x_{n_k})$ is $0$ for even $k$ and $1$ for odd $k$ (if you follow the convention $0\notin \mathbb{N}$, switch even and odd), so $(x_{n_k})$ is not convergent.
If $E$ is a normed space, then the closed unit ball of $E^\ast$ is compact in the weak$^\ast$ topology by the Banach-Alaoglu theorem, and under some conditions on $E$ it is also sequentially compact.
- If $E$ is separable, then the subspace topology induced on the closed unit ball of $E^\ast$ by the weak$^\ast$ topology is metrisable (Note: The weak$^\ast$ topology on $E^\ast$ is then generally not metrisable itself), hence the closed unit ball of $E^\ast$ is then weak$^\ast$-sequentially compact.
- If $E$ is reflexive, the closed unit ball of $E^\ast$ is weak$^\ast$-sequentially compact.
$\ell^\infty$ is neither separable nor reflexive.
Best Answer
Your proof is correct. In fact, it generalizes to the case that $F_1$ is a metrizable subset of an arbitrary space $X$.