Proof.

Bounded

Fix any \(x_0\in X\). The open balls \(B(x_0;n)\), for \(n=1,2,3,\dots \), cover the whole of \(X\) and in particular cover \(K\). By compactness finitely many of them cover \(K\), and since the balls increase with \(n\) the largest of that finite collection contains the rest: \(K\subset B(x_0;N)\) for some \(N\). Hence \(K\) is bounded.

Closed

We show the complement is open. Let \(x\in X\setminus K\). For each \(y\in K\) put \[r_y = \tfrac 12\, d(x,y) > 0 ,\] which is positive precisely because \(x\notin K\) and so \(x \neq y\). The balls \(B(y;r_y)\), as \(y\) runs over \(K\), form an open cover of \(K\), so finitely many of them cover it, say \(B(y_1;r_1),\dots ,B(y_n;r_n)\) with \(r_i = r_{y_i}\). Put \[r = \min \{r_1,\dots ,r_n\} > 0 .\] Then \(B(x;r)\) meets none of these balls. For if \(z\in B(x;r)\cap B(y_i;r_i)\) then \[d(x,y_i)\leq d(x,z)+d(z,y_i) < r + r_i \leq 2r_i = d(x,y_i),\] which is absurd. So \(B(x;r)\) misses \(K\) entirely, that is \(B(x;r)\subset X\setminus K\). As \(x\) was arbitrary, \(X\setminus K\) is open and \(K\) is closed. □

Proof. Since \(E\) is bounded, we have \(E\subset I_1=[a,b]\), \((a,b)\). We proceed by contradiction on \(I_1\).

Suppose (if possible) there is an open covering \(\{G_i\}\) of \(I_1\) i.e \(I_1\subset \bigcup G_i\) which has no finite sub covering. Bisect \(I_1\) into two closed intervals \(I_2\) and \(I'_2\). At least one of these must have no finite sub covering. Let \(I_2\) be the part without a finite sub covering. Bisect \(I_2\) into \(I_3\) and \(I'_3\). Continuing in this way, we have a nested sequence of intervals \(I_n\) such that \(I_n\) has no finite subcover for each \(n\).

By nested sequence interval theorem \(\displaystyle {\bigcap _{n=1}^{\infty }I_n\neq \emptyset }.\) Let \(x\in I_n\). Then \(\displaystyle { x\in I_1\subset \bigcup ^{\infty }_{i=1}G_i},\hspace {0.2cm}\) so, \(x\in G_i\) for some \(i\). Since \(G_i\) is open there exist an open ball \(B_r(x)=(x-r,x+r)\subset G_i.\) Taking \(n\) large enough, we have \(I_n\subset B_r(x)=(x-r,x+r)\subset G_i.\) This means that \(I_n\) is covered by a single set \(G_i\). This contradicts that \(I_n\) has no finite sub covering. Thus our early assumption is not true. Therefore \(E\) has a finite sub covering. □

Proof. One direction is the theorem above, applied coordinatewise: a closed bounded subset of \(\mathbb {R}^n\) lies in a closed box \([a_1,b_1]\times \cdots \times [a_n,b_n]\), which is compact by repeating the bisection argument in \(n\) dimensions, and a closed subset of a compact set is compact. The other direction is the previous theorem, which holds in any metric space. □

Solution. Consider \(G_n=\Big (a-\dfrac {1}{n},b+\dfrac {1}{n}\Big )\hspace {0.2cm}\) then \(\displaystyle {[a,b]\subset \bigcup ^{\infty }_{n=1}G_n}\hspace {0.2cm}\) and that \([a,b]\subset (a-1,b+1)=G_1,\hspace {0.2cm}\) thus \(\{G_1\}\) is a finite sub cover of \([a,b]\).

Solution. Take \(\displaystyle {G_n = \Big (\frac 1n , 2\Big )}\) for \(n=1,2,3,\dots \) Each is open, and every \(x\in (0,1]\) satisfies \(x>\frac 1n\) for some \(n\), so \(\{G_n\}\) is an open cover. Any finite subfamily has a largest index \(N\), and the union of the subfamily is then \(\Big (\dfrac 1N,2\Big )\), which omits \(\dfrac {1}{N+1}\). No finite subcover exists.

The set is bounded but not closed, and the cover exhibits exactly the point that is missing. This is the standard way to prove non-compactness: produce one cover whose finite subfamilies visibly fail.