We wish to show that intervals (with standard topology) are connected. In order to this, we will prove that the space of real numbers $\mathbb{R}$ is connected. First we need a lemma.

Let $(X,d)$ be a metric space. Recall that for $x\in X$ and $r\in\mathbb{R}^{+}$ we have $$B(x,r)=\{y\in X\ |\ d(x,y)<r\}.$$

Lemma. Let $(X,d)$ be a metric space and $R\subset\mathbb{R}^{+}$ such that $R$ is nonempty and bounded. Then for any $x\in X$ we have $$\bigcup_{r\in R}B(x,r)=B(x,\mathrm{sup}(R)).$$

Proof. Assume that $y\in\bigcup_{r\in R}B(x,r)$ Then there is $r_0\in R$ such that $d(x,y)<r_0$ and thus $d(x,y)<\mathrm{sup}(R)$ so $y\in B(x,\mathrm{sup}(R))$

Now assume that $y\in B(x,\mathrm{sup}(R))$ Then $d(x,y)<\mathrm{sup}(R)$ and it follows (from the definition of supremum) that there is $r_0\in R$ such that $d(x,y)<r_0$ and therefore $y\in B(x,r_0)\subset\bigcup_{r\in R}B(x,r)$ which completes the proof. $\square$

Proposition. The space of real numbers is connected.

Proof. Assume that $U,V \subseteq\mathbb{R}$ are open subsets of $\mathbb{R}$ such that $U\cap V = \emptyset$ and $U\cup V = \mathbb{R}$ Furthermore assume that $U \neq \emptyset$ and take any $x_{0} \in U$ Then (since $U$ is open) there is $r_{0}\in\mathbb{R}$ such that the open ball $$B(x_{0},r_{0})=\{ x\in \mathbb{R}\ |\ |x-x_{0}|<r_{0}\} $$ is contained in $U$ Consider the set $$R=\{ r\in\mathbb{R}^{+}\ |\ B(x_0,r)\subseteq U\} .$$ Thus $R$ is nonempty.

Assume that $R$ is bounded. Denote by $s=\mathrm{sup}(R)<\infty$ We can apply the lemma: $$\bigcup_{r\in R} B(x_0,r)=B(x_0,s).$$ Thus (due to the definition of $R$ $B(x_0,s)$ is a maximal open ball (with the center in $x_0$ which is contained in $U$ Now $$B(x_0,s)=(a,b)$$ for some $a,b\in\mathbb{R}$ Since $(a,b)$ is maximal then $a\not\in U$ or $b\not\in U$ Indeed, if both $a\in U$ and $b\in U$ then (since $U$ is open) small neighbourhoods of $a$ and $b$ are also contained in $U$ so $(a-\epsilon,b+\epsilon)$ is contained in $U$ (for some $\epsilon>0$ , but $(a,b)$ was maximal. Contradiction.

Without loss of generality we can assume that $b\not\in U$ Then $b\in V$ because $U\cup V=\mathbb{R}$ But then (since $V$ is open) there is $c\in\mathbb{R}$ such that $a<c<b$ and $c\in V$ Thus $U\cap V\neq\emptyset$ Contradiction. Therefore $R$ is unbounded.

Take any unbounded sequence $(a_n)_{n=1}^{\infty}$ from $R$ Then we have $$\mathbb{R}=\bigcup_{n=1}^{\infty} B(x_0,a_n)\subseteq U$$ and thus $U=\mathbb{R}$ so $V=\emptyset$ This completes the proof. $\square$ Corollary. For any $a,b\in\mathbb{R}$ such that $a<b$ intervals $(a,b)$ $[a,b)$ $(a,b]$ and $[a,b]$ are connected.

Proof. One can easily show that intervals are continous image of $\mathbb{R}$ and therefore intervals are connected.