[proofplan] We prove the result by contradiction. The [discrete metric](/page/Discrete%20Metric) makes every singleton in $X$ open, so if a path $\gamma$ took two distinct values, the preimages of one chosen value and its complement would separate the interval $[0,1]$. We then prove directly that $[0,1]$ admits no separation, using the least upper bound property of $\mathbb{R}$. [/proofplan] [step:Show that singletons are open in the discrete metric topology] Let $x\in X$ be arbitrary. Define the open metric ball $B_d(x,1/2)\subset X$ by \begin{align*}B_d(x,1/2)=\{y\in X:d(x,y)<1/2\}.\end{align*} If $y\in B_d(x,1/2)$, then $d(x,y)<1/2<1$. By the definition of the discrete metric, $d(x,y)=1$ whenever $x\neq y$, so $y=x$. Conversely, $d(x,x)=0<1/2$, so $x\in B_d(x,1/2)$. Hence \begin{align*}B_d(x,1/2)=\{x\}.\end{align*} Thus every singleton $\{x\}$ is open in the metric topology induced by $d$. [guided] The only special property of the discrete metric that we need is that each point is isolated. Fix a point $x\in X$. The metric ball of radius $1/2$ around $x$ is \begin{align*} B_d(x,1/2)=\{y\in X:d(x,y)<1/2\}. \end{align*} Why choose radius $1/2$? Because the discrete metric has only two possible values: $0$ and $1$. If $y=x$, then $d(x,y)=d(x,x)=0<1/2$, so $y$ lies in the ball. If $y\neq x$, then $d(x,y)=1$, and $1$ is not less than $1/2$, so $y$ does not lie in the ball. Therefore \begin{align*}B_d(x,1/2)=\{x\}.\end{align*} Since metric balls are open by definition of the metric topology, the singleton $\{x\}$ is open. Because $x\in X$ was arbitrary, every singleton in $X$ is open. [/guided] [/step] [step:Prove that the interval $[0,1]$ has no separation] We first record the elementary connectedness fact needed for the proof. By a separation of $[0,1]$ we mean a pair of subsets $U,V\subset[0,1]$ such that $U$ and $V$ are nonempty, open in the [subspace topology](/page/Subspace%20Topology) on $[0,1]$, disjoint, and satisfy $U\cup V=[0,1]$. [claim:The interval $[0,1]$ admits no separation] There is no separation of $[0,1]$. [/claim] [proof] Assume, toward a contradiction, that $U,V\subset[0,1]$ form a separation of $[0,1]$. Since $0\in[0,1]=U\cup V$, after interchanging the names of $U$ and $V$ if necessary, assume $0\in U$. Define the set \begin{align*}S=\{t\in[0,1]:[0,t]\subset U\}.\end{align*} The set $S$ is nonempty because $0\in U$, so $0\in S$. It is bounded above by $1$, so the least upper bound property of $\mathbb{R}$ gives a number $a=\sup S\in[0,1]$. We claim that $a\in U$. Suppose instead that $a\in V$. Since $V$ is open in the subspace topology on $[0,1]$, there exists $\varepsilon>0$ such that \begin{align*}(a-\varepsilon,a+\varepsilon)\cap[0,1]\subset V.\end{align*} Because $a=\sup S$, there exists $s\in S$ with $a-\varepsilon/2<s\le a$, unless $a=0$. If $a=0$, then $0\in U\cap V$, contradicting $U\cap V=\varnothing$. Thus we may take such an $s$. Since $s\in S$, we have $[0,s]\subset U$, so $s\in U$. But $s\in(a-\varepsilon,a+\varepsilon)\cap[0,1]\subset V$, contradicting $U\cap V=\varnothing$. Therefore $a\in U$. Since $U$ is open in the subspace topology on $[0,1]$, there exists $\delta>0$ such that \begin{align*}(a-\delta,a+\delta)\cap[0,1]\subset U.\end{align*} Assume first that $a<1$. Define $r\in[0,1]$ by \begin{align*}r=a+\frac{1}{2}\min\{\delta,1-a\}.\end{align*} Then $a<r<1$ and $r<a+\delta$. Since $a=\sup S$ and $a-\delta<a$, the number $a-\delta$ is not an upper bound for $S$. Hence there exists $s\in S$ such that \begin{align*}a-\delta<s\leq a.\end{align*} By the definition of $S$, the inclusion $[0,s]\subset U$ holds. If $q\in[0,r]$, then either $q\leq s$, in which case $q\in U$, or $s<q\leq r$, in which case \begin{align*} a-\delta<s<q\leq r<a+\delta, \end{align*} so $q\in(a-\delta,a+\delta)\cap[0,1]\subset U$. Therefore $[0,r]\subset U$, so $r\in S$. This contradicts that $a$ is an upper bound for $S$. Hence $a=1$. It remains to prove that $U=[0,1]$ without assuming the conclusion. We already know $1=a\in U$. Let $t\in[0,1)$ be arbitrary. Since $a=\sup S=1$ and $t<1$, the number $t$ is not an upper bound for $S$. Hence there exists $s\in S$ such that $t<s\leq1$. By the definition of $S$, $[0,s]\subset U$, and therefore $t\in U$. Thus every $t\in[0,1)$ belongs to $U$, and $1\in U$, so $[0,1]\subset U$. Hence $U=[0,1]$, contradicting that $V$ is nonempty. Therefore no separation of $[0,1]$ exists. [/proof] [/step] [step:Use continuity to turn two path values into a separation of $[0,1]$] Let \begin{align*}\gamma:[0,1]\to X\end{align*} be continuous. Define $x_0=\gamma(0)\in X$. We prove that $\gamma(t)=x_0$ for every $t\in[0,1]$. Assume, toward a contradiction, that there exists $t_1\in[0,1]$ such that $\gamma(t_1)\neq x_0$. Define \begin{align*}A=\gamma^{-1}(\{x_0\})\subset[0,1]\end{align*} and \begin{align*}C=\gamma^{-1}(X\setminus\{x_0\})\subset[0,1].\end{align*} The set $\{x_0\}$ is open in $X$ by the previous step. Also $X\setminus\{x_0\}$ is open: for each $y\in X\setminus\{x_0\}$, the ball $B_d(y,1/2)=\{y\}$ is contained in $X\setminus\{x_0\}$. Since $\gamma$ is continuous, both $A$ and $C$ are open in the subspace topology on $[0,1]$. Moreover, $0\in A$, because $\gamma(0)=x_0$, and $t_1\in C$, because $\gamma(t_1)\neq x_0$. Thus $A$ and $C$ are nonempty. They are disjoint by definition, and \begin{align*} A\cup C=[0,1], \end{align*} because every value $\gamma(t)$ either equals $x_0$ or lies in $X\setminus\{x_0\}$. Hence $A$ and $C$ form a separation of $[0,1]$, contradicting the preceding step. [/step] [step:Conclude that the path is constant] The contradiction shows that there is no $t_1\in[0,1]$ with $\gamma(t_1)\neq x_0$. Therefore \begin{align*}\gamma(t)=x_0\end{align*} for every $t\in[0,1]$. Since $x_0=\gamma(0)\in X$, this proves that there exists $x_0\in X$ such that $\gamma(t)=x_0$ for all $t\in[0,1]$. [/step]