bin:no-ansi-sgr: Set LC_ALL=C for [0-?] range

Setting LC_ALL=C makes it possible to use the range `[0-?]` instead of
splitting it into `[0-9:-?]` as done previously. Without LC_ALL, sed
complains with:

	sed: -e expression #1, char 21: Invalid range end

The GNU manual explains this partially, although I still don't quite
understand why this range specifically does not work in `en_US.utf8`.
See:

> Within a bracket expression, a *range expression* consists of two
> characters separated by a hyphen. It matches any single character that
> sorts between the two characters, inclusive. In the default C locale,
> the sorting sequence is the native character order; for example,
> `[a-d]` is equivalent to `[abcd]`.

Link: https://www.gnu.org/software/sed/manual/sed.html#Character-Classes-and-Bracket-Expressions

.local/bin/no-ansi-sgr

@@ -1,9 +1,8 @@
-#!/bin/sed -f
+#!/bin/sh
 # SPDX-License-Identifier: MIT
 # Copyright (c) 2025 Julian Prein
 #
 # Remove ANSI SGR (Select Graphic Rendition) escape sequences, e.g. setting
 # color or bold font.
 
-# TODO: Why is the range `[0-?]` invalid?
-s/\[[0-9:-?]*[ -/]*m//g
+env LC_ALL=C sed 's/\[[0-?]*[ -/]*m//g'