FALSE: a function with the intermediate value property on an interval is continuous

Define $\psi :\mathbb{R}\to \mathbb{R}$ by $\psi (u):=\min \{u-\lfloor u\rfloor ,\lfloor u\rfloor +1-u\}$, the distance from $u$ to the nearest integer, and define $g:[-1,1]\to \mathbb{R}$ by $g(x):=\psi (1\mathrm{/}x)$ for $x\mathrm{\ne }0$ and $g(0):=0$.

$0\le \psi (u)\le 1\mathrm{/}2$ for every real $u$: writing $\theta :=u-\lfloor u\rfloor \in [0,1)$, the two entries are $\theta \ge 0$ and $1-\theta >0$, and their minimum is at most their average $1\mathrm{/}2$. Consequently $0\le g(x)\le 1\mathrm{/}2$ for every $x\in [-1,1]$.

$\psi (u)=\min \{\mathrm{\mid }u-m\mathrm{\mid }:m\in \mathbb{Z}\}$ in the sense that $\mathrm{\mid }u-m\mathrm{\mid }\ge \psi (u)$ for every integer $m$, with equality for $m=\lfloor u\rfloor$ or $m=\lfloor u\rfloor +1$. Indeed, with $n:=\lfloor u\rfloor$: for $m\le n$ one has $\mathrm{\mid }u-m\mathrm{\mid }=u-m\ge u-n$, and for $m\ge n+1$ one has $\mathrm{\mid }u-m\mathrm{\mid }=m-u\ge n+1-u$.

For every real $\eta >0$ and every $y\in [0,1\mathrm{/}2]$ there is $x$ with $0<x<\eta$ and $g(x)=y$: take a natural $k\ge 1$ with $1\mathrm{/}\iota (k)<\eta$ and put $x:=1\mathrm{/}(\iota (k)+y)$. Then $0<x\le 1\mathrm{/}\iota (k)<\eta$ and $1\mathrm{/}x=\iota (k)+y$ with $0\le y\le 1\mathrm{/}2<1$, so $\lfloor 1\mathrm{/}x\rfloor =\iota (k)$ and $\psi (1\mathrm{/}x)=\min \{y,1-y\}=y$.

$\psi$ is continuous on $\mathbb{R}$, because $\mathrm{\mid }\psi (u)-\psi (v)\mathrm{\mid }\le \mathrm{\mid }u-v\mathrm{\mid }$ for all reals $u,v$: choose an integer $m$ with $\mathrm{\mid }u-m\mathrm{\mid }=\psi (u)$, which exists by step 2.2; then $\psi (v)\le \mathrm{\mid }v-m\mathrm{\mid }\le \mathrm{\mid }v-u\mathrm{\mid }+\mathrm{\mid }u-m\mathrm{\mid }=\mathrm{\mid }u-v\mathrm{\mid }+\psi (u)$, and exchanging $u$ and $v$ gives the other inequality. So $\delta :=\epsilon$ witnesses continuity at every point.

For every real $\eta >0$ and every $y\in [0,1\mathrm{/}2]$ there is $x$ with $-\eta <x<0$ and $g(x)=y$: with $k$ as in step 2.3 put $x:=-1\mathrm{/}(\iota (k)+y)$, so $-\eta <x<0$ and $1\mathrm{/}x=-\iota (k)-y$. If $y=0$ then $1\mathrm{/}x=-\iota (k)$ is an integer and $\psi (1\mathrm{/}x)=0=y$; if $0<y\le 1\mathrm{/}2$ then $\lfloor 1\mathrm{/}x\rfloor =-\iota (k)-1$ and $1\mathrm{/}x-\lfloor 1\mathrm{/}x\rfloor =1-y$, so $\psi (1\mathrm{/}x)=\min \{1-y,y\}=y$.

$g$ is discontinuous at $0$: $g(0)=0$, and by step 2.3 every real $\delta >0$ admits $x$ with $0<x<\delta$ and $g(x)=1\mathrm{/}2$, so $\mathrm{\mid }g(x)-g(0)\mathrm{\mid }=1\mathrm{/}2$. Hence no $\delta$ witnesses the continuity condition at $0$ for $\epsilon :=1\mathrm{/}2$.

$g$ is continuous at every $x\in [-1,1]$ with $x\mathrm{\ne }0$: on the set $\{x\in [-1,1]:x\mathrm{\ne }0\}$ the map $x\mapsto 1\mathrm{/}x$ is continuous, and $g$ is its composite with $\psi$; continuity at a point of that set is continuity of $g$ there, since the set contains a whole neighbourhood of $x$ inside $[-1,1]$ when $x\mathrm{\ne }0$.

If $0\in [a,b]$ then, since $a<b$, either $b>0$ or $a<0$. In the first case step 2.3 with $\eta :=b$ gives $c$ with $0<c<b$ and $g(c)=y$, and $c\in [a,b]$; in the second case step 3.2 with $\eta :=-a$ gives $c$ with $a<c<0$ and $g(c)=y$. Either way $y\in g[[a,b]]$.

$g$ has the intermediate value property on $[-1,1]$. Let $a<b$ in $[-1,1]$ and let $y$ lie between $g(a)$ and $g(b)$ in either order; in particular $y\in [0,1\mathrm{/}2]$ by step 2.1. If $0\mathrm{\notin }[a,b]$ then $g$ restricted to $[a,b]$ is continuous by step 4.1, and the intermediate value theorem supplies $c\in [a,b]$ with $g(c)=y$.

So $g$ is a function on the interval $[-1,1]$ with the intermediate value property that is not continuous on $[-1,1]$, and the claim in the Statement is false.