Proof: Suppose $\O$ is open and $x\in\O.$ Since $\O$ is open and non-empty, it is a union of open intervals. Since $x\in\O,$ $x$ is an element of one of those intervals. That interval must contain $(x-\eps,x+\eps)$ for some $\eps>0.$ (Just take $\eps$ to be smaller than the distance between $x$ and the closest endpoint of the interval.) Since that interval is a subset of $\O,$ it follows that $(x-\eps,x+\eps)\subseteq\O.$

Conversely, suppose that for every $x\in\O,$ there is an $\eps_x>0$ such that $(x-\eps_x,x+\eps_x)\subseteq\O.$ Then $\O=\bigcup_{x\in\O}(x-\eps_x,e+\eps).$ (The union is contained in $\O$ because each set in the union is a subset of $\O$; the union contains $\O$ because every $x\in\O$ is an element of one of the sets in the union, namely of $(x-\eps_x,x+\eps_x).$) Since $(x-\eps_x,x+\eps_x)$ is an open interval, we have written $\O$ as a union of open intervals. $\qed$

Proof: Let ${\mathscr C}=\{\O_\alpha\,|\,\alpha\in A\}$ be an open cover of $[a,b].$ Note that for any $c\in[a,b],$ $\mathscr C$ is an open cover of $[a,c].$ Define $X=\{c\in[a,b]\,|\,[a,c]$ has a finite subcover for $\mathscr C\}.$ Since $a$ is contained in some $\O_\alpha,$ $[a,a]$ has a subcover consisting of a single set. So $a\in X,$ and $X$ is non-empty. Since $X\subseteq [a,b],$ $X$ is bounded. Therefore, as a bounded and non-empty set, $X$ has a least upper bound by the least upper bound axiom. Let $\lambda=lub(X).$ We will be done if we can show $\lambda\in X$ and $\lambda=b,$ since that will mean by definition of $X$ that $[a,b]$ has a finite subcover for ${\mathscr C}.$

We first show that $\lambda\in X.$ Because $b$ is an upper bound for $X,$ we know that $\lambda\le b,$ so $\lambda\in[a,b].$ Since $\mathscr C$ is an open cover of $[a,b],$ there is a $\beta\in A$ such that $\lambda\in\O_\beta.$ Since $\O_\beta$ is open, there is an $\eps>0$ such that $(\lambda-\eps,\lambda+\eps)\subset\O_\beta.$ Since $\lambda=lub(X),$ there is an $x\in X$ such that $x>\lambda-\eps,$ by the theorem at the end of the third reading guide. Note that $x\in\O_\beta.$ Since $x\in X,$ then by defintion of $X,$ there is a finite subcover $\{\O_{\alpha_1},\O_{\alpha_2},\dots,\O_{\alpha_k}\}$ of $[a,x]$ from $\mathscr C.$ But then, adding $\O_\beta$ to that finite subcover, we get that $\{\O_\beta,\O_{\alpha_1},\O_{\alpha_2},\dots,\O_{\alpha_k}\}$ is a finite subcover of $[a,\lambda].$ And by definition of $X,$ this means that $\lambda\in X.$

Finally, we must show that $\lambda=b.$ We know $\lambda\le b,$ so suppose, for the sake of contradiction, that $\lambda<b.$ Using the same $\O_\beta$ and $\eps$ as in the preceding paragraph, Let $y=x+\frac12\min(\eps,b-\lambda).$ Then $\lambda<y<b,$ and $\{\O_\beta,\O_{\alpha_1},\O_{\alpha_2},\dots,\O_{\alpha_k}\}$ is a finite subcover of $[a,y]$ from $\mathscr C.$ But then $y\in X$ by definition of $X,$ and the fact that $y>\lambda$ contradicts the fact that $\lambda$ is an upper bound for $X.$

Proof: Let $X$ be a bounded, infinte subset of $X.$ Suppose, for the sake of contradiction that $X$ does not have an accumulation point. That is, for every $c\in\R,$ there is some $\eps_c>0$ such that no point of $X,$ other than possibly $c$ itself, is within distance $\eps$ of $c.$ Another way of saying this is that the open interval $(c-\eps_c,c+\eps_c)$ is either empty (in the case $c\notin X$) or contains just the single point $c$ (in the case $c\in X$).

Let $a$ be a lower bound for $X,$ and let $b$ be an upper bound for $X,$ so that $X\subseteq[a,b].$ We construct an open cover of $[a,b].$ Suppose $c\in[a,b].$ Let $\eps_c$ be as in the preceding paragraph, and let $\O_c=(c-\eps_c,c+\eps_c).$ Then $\{\O_c\,|\,c\in[a,b]\}$ is an open cover of $[a,b].$ By the Heine-Borel Theorem, there is a finite subcover, $\{\O_{c_1},\O_{c_2},\dots,\O_{c_k}\}.$ So, we have $X\subseteq[a,b]\subseteq\bigcup_{i=1}^k\O_{c_i}.$ But each of the sets $\O_{c_i}$ contains at most one point of $X,$ so $X$ can have no more than $k$ elements. $\qed$