Every argument that I can see right now to show that (5) implies (4) either essentially goes through one of the other equivalent forms or uses a much more sophisticated result about metric spaces, namely, that every metric space is paracompact. This means that every open cover $\mathscr{U}$ of $X$ has a locally finite open refinement $\mathscr{V}$ covering $X$. That is,
- $\mathscr{V}$ is an open cover of $X$;
- for each $V\in\mathscr{V}$ there is a $U\in\mathscr{U}$ such that $V\subseteq U$; and
- each $x\in X$ has an open nbhd $N_x$ such that $\{V\in\mathscr{V}:N_x\cap V\ne\varnothing\}$ is finite.
Note that the third condition implies that each point of $X$ is in only finitely many members of $\mathscr{V}$, i.e., that $\mathscr{V}$ is point-finite. This is actually all that I need. (A space in which every open cover has a point-finite open refinement is said to be metacompact, so I’m actually using only the weaker result that every metric space is metacompact.)
Theorem: Every point-finite open cover of $X$ has an irreducible subcover, meaning one with no proper subcover.
Proof: Let $\mathfrak{R}=\{\mathscr{R}\subseteq\mathscr{V}:\mathscr{R}\text{ covers }X\}$; $\mathfrak{R}$ is partially ordered by $\supseteq$. Let $\mathfrak{C}$ be a chain in $\mathfrak{R}$, and let $\mathscr{C}=\bigcap\mathfrak{C}$; I claim that $\mathscr{C}\in\mathfrak{R}$, i.e., that $\mathscr{C}$ still covers $X$.
Proof of Claim: Suppose that some $x\in X$ is not covered by $\mathscr{C}$. Let $V_1,\dots,V_n$ be the finitely many members of $\mathscr{V}$ containing $x$. Then none of these $V_k$ can belong to $\mathscr{C}$ (or else $x$ would be covered by $\mathscr{C}$). But $\mathscr{C}$ is the intersection of the collections in the chain $\mathfrak{C}$, so for each $k=1,\dots,n$ there is some $\mathscr{C}_k\in\mathfrak{C}$ such that $V_k\notin\mathscr{C}_k$. Because $\mathfrak{C}$ is a chain, the collections $\mathscr{C}_1,\dots,\mathscr{C}_n$ are nested, and without loss of generality we may assume that the indexing has been chosen so that $\mathscr{C}_1\supseteq\dots\supseteq\mathscr{C}_n$. But then $\mathscr{C}_n$ contains none of the sets $V_1,\dots,V_n$, so $\mathscr{C}_n$ does not cover $x$, and hence $\mathscr{C}_n\notin\mathfrak{R}$, a contradiction.
We can now apply Zorn’s lemma to the partial order $\langle\mathfrak{R},\supseteq\rangle$ to conclude that $\mathfrak{R}$ has a maximal element $\mathscr{M}$ with respect to $\supseteq$: that is, $\mathscr{M}$ is in $\mathfrak{R}$, but no proper subcollection of $\mathscr{M}$ belongs to $\mathfrak{R}$. But then $\mathscr{M}$ is an open cover of $X$ with no proper subcover, i.e., an irreducible cover of $X$.$\dashv$
Now it’s easy to show that (5) implies (4). Suppose that every infinite open cover of $X$ has a proper subcover; this amounts to saying that every irreducible open cover of $X$ is finite. Let $\mathscr{U}$ be an open cover of $X$. By what we just showed, $\mathscr{U}$ has an irreducible subcover $\mathscr{V}$, and being irreducible, $\mathscr{V}$ must be finite. Thus, $X$ is compact.
Best Answer
This is true but not very easy to prove.
Suppose $X$ is not compact. Without loss of generality assume that the original metric $d$ on $X$ is such that $d(x,y)<1$ for all $x,y\in X.$ There exists a decreasing sequence of non-empty closed sets $\{C_{n}\}$ whose intersection is empty. Let $$\rho (x,y)=\sum_{n=1}^{\infty }\frac{1}{% 2^{n}}d_{n}(x,y)$$ where $$d_{n}(x,y)=\left\vert d(x,C_{n})-d(y,C_{n})\right\vert +\min \{d(x,C_{n}),d(y,C_{n})\}d(x,y).$$ We claim that $\rho $ is a metric on $X$ which is equivalent to $d$ and that $% (X,\rho )$ is not complete. Note that $d_{n}(x,y)\leq 2$ for all $x,y\in X.$ If $x$ and $y$ $\in C_{k}$ then $x$ and $y$ $\in C_{n}$ for $1\leq n\leq k$ and hence $\rho (x,y)\leq \sum_{n=k+1}^{\infty }\frac{2}{2^{n}}=% \frac{1}{2^{k}}$. Thus, the diameter of $C_{k}$ in $(X,\rho )$ does not exceed $\frac{1}{2^{k}}$. Once we prove that $\rho $ is a metric equivalent to $d$ it follows that $\rho $ is not complete because $\{C_{n}\}$ is a decreasing sequence of non-empty closed sets whose intersection is empty.
Assuming (for the time being) that $d_{n}$ satisfies triangle inequality it follows easily that $\rho $ is a metric: if $\rho (x,y)=0$ then $% d(x,C_{n})=d(y,C_{n})$ for each $n$ and $\min \{d(x,C_{n}),d(y,C_{n})\}d(x,y)=0$ for each $n$. If $d(x,y)\neq 0$ it follows that $d(x,C_{n})=d(y,C_{n})=0$ for each $n$ which implies that $x$ and $y$ belong to each $C_{n}$ contradicting the hypothesis. Thus $\rho $\ is a metric. Also $\rho (x_{j},x)\rightarrow 0$ as $j\rightarrow \infty $ implies $\left\vert d(x_{j},C_{n})-d(x,C_{n})\right\vert \rightarrow 0$ and $% \min \{d(x_{j},C_{n}),d(x,C_{n})\}d(x_{j},x)\rightarrow 0$ as $j\rightarrow \infty $ for each $n$. There is at least one integer $k$ such that $x\notin C_{k}$ and we conclude that $d(x_{j},x)\rightarrow 0$. Conversely, suppose $% d(x_{j},x)\rightarrow 0$. Then $d_{n}(x_{j},x)\rightarrow 0$ for each $n$ and the series defining $\rho $ is uniformly convergent, so $\rho (x_{j},x)\rightarrow 0$. It remains only to show that $d_{n}$ satisfies triangle inequality for each $n$ . We have to show that $$\left\vert d(x,C_{n})-d(y,C_{n})\right\vert$$ $$+\min \{d(x,C_{n}),d(y,C_{n})\}d(x,y)$$
$$\leq \left\vert d(x,C_{n})-d(z,C_{n})\right\vert +\min \{d(x,C_{n}),d(z,C_{n})\}d(x,z)$$ $$+\left\vert d(z,C_{n})-d(y,C_{n})\right\vert +\min \{d(z,C_{n}),d(y,C_{n})\}d(z,y)$$ for all $x,y,z.$ Let $% r_{1}=d(x,C_{n}),r_{2}=d(y,C_{n}),r_{3}=d(z,C_{n})$. We consider six cases depending on the way the numbers $r_{1},r_{2},r_{3}$ are ordered. It turns out that the proof is easy when $r_{1}$ or $r_{2}$ is the smallest of the three. We give the proof for the case $r_{3}\leq r_{1}\leq r_{2}$. (The case $r_{3}\leq r_{2}\leq r_{1}$ is similar). We have to show that
$$r_{2}-r_{1}+r_{1}d(x,y)\leq r_{1}-r_{3}+r_{3}d(x,z)+r_{2}-r_{3}+r_{3}d(z,y)$$ which says $$r_{1}d(x,y)\leq 2r_{1}-2r_{3}+r_{3}d(x,z)+r_{3}d(z,y).$$ Since $d$ satisfies trangle inequality it suffices to show that $$% r_{1}d(x,z)+r_{1}d(z,y)\leq 2r_{1}-2r_{3}+r_{3}d(x,z)+r_{3}d(z,y).$$ But this last inequality is equivalent to $$(r_{1}-r_{3})[d(x,z)+d(z,y)]\leq 2r_{1}-2r_{3}.$$ This is true because $d(x,z)+d(z,y)\leq 1+1=2$.