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 $g(x) := \psi(1/x)$ for $x \ne 0$ and $g(0) := 0$.

$0 \le \psi(u) \le 1/2$ 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 $1/2$. Consequently $0 \le g(x) \le 1/2$ for every $x \in [-1,1]$.

$\psi(u) = \min\{\, |u - m| : m \in \mathbb{Z} \,\}$ in the sense that $|u - m| \ge \psi(u)$ 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 $|u - m| = u - m \ge u - n$, and for $m \ge n+1$ one has $|u - m| = m - u \ge n + 1 - u$.

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

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

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

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

$g$ is continuous at every $x \in [-1,1]$ with $x \ne 0$: on the set $\{x \in [-1,1] : x \ne 0\}$ the map $x \mapsto 1/x$ 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 $x \ne 0$.

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 $y \in [0,1/2]$ by step 2.1. If $0 \notin [a,b]$ 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.