= Solution
Use the <product topology> on $\mathcal C=[k]^{\mathbb Z}$, with $[k]$ discrete, and the <left shift> $(\mathcal Lc)(j)=c(j+1)$. A basic <cylinder set> specifies finitely many coordinates. This is a <compact metric space>, and $\mathcal L$ is a <homeomorphism>. Write $Y=\overline{\{\mathcal L^n c:n\geq0\}}$ for the forward <orbit closure>. A <minimal point> means that $Y$ is a <minimal dynamical system>, equivalently that every point of $Y$ has a dense forward <orbit> in $Y$; it need not be a <fixed point>.
There is a convention needed in the source: the bounded-gaps property concerns every \b[finite] <integer interval> $I$, hence every finite <word over an alphabet>. Literally allowing $I=\mathbb Z$ would make the displayed condition impossible for finite $U$, although a constant coloring is a <minimal point>. Under the standard finite-word interpretation, the property is precisely <uniform recurrence>.
Suppose first that $Y$ is a <minimal dynamical system>. Its nonempty compact forward-invariant <subset> $\mathcal L(Y)$ must equal $Y$. As the ambient <left shift> is invertible, all <integer> translates of $c$ belong to $Y$. Let $I=[a,b]\cap\mathbb Z$ have length $\ell=b-a+1$, and let $V=\{z\in Y:z|_I=c|_I\}$. This is a nonempty <clopen set>. Each forward <orbit> in $Y$ meets $V$, so $\{\mathcal L^{-n}V:n\geq0\}$ covers $Y$. By <compactness>, finitely many suffice; let $D$ be the largest index in this finite cover. For any <integer> $u$, the point $\mathcal L^{u-a}c$ lies in $Y$, so some $0\leq n\leq D$ satisfies $\mathcal L^{u-a+n}c\in V$. The prescribed word therefore occurs at positions $[u+n,u+n+\ell-1]$, entirely inside $[u,u+D+\ell-1]$. Taking $M=D+\ell$ proves the bounded-gaps property for every interval of length at least $M$.
Conversely suppose $c$ is <uniformly recurrent>. Every finite word from any <integer> translate of $c$ occurs arbitrarily far to the right, so every such translate belongs to $Y$. Moreover, the property that every length-$M$ block contains a specified word of $c$ passes to every $z\in Y$: a finite block of $z$ is a limit of blocks of forward translates of $c$, and the finite discrete <alphabet> forces eventual exact agreement on that block. Given $I=[a,b]\cap\mathbb Z$, look for its word inside $z([a,a+M-1])$. An occurrence starts at $a+n$ for some $n\geq0$, so $\mathcal L^n z$ agrees with $c$ on $I$. Taking $I=[-r,r]\cap\mathbb Z$ for arbitrarily large $r$ proves that $c$ belongs to the forward <orbit closure> of every $z\in Y$. That <closure> is a closed forward-invariant <subset> of $Y$, so it also contains every forward translate of $c$ and hence all of $Y$. Every forward <orbit> is therefore dense in $Y$, proving
$$
\boxed{c\text{ is a minimal point}\iff c\text{ is uniformly recurrent}.}
$$
The same conclusion holds if <orbit closure> is defined using all <integer> iterates: the bounded-gaps condition makes the forward and two-sided closures equal.
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