 Solution

For a strict-range modification of a continuous extension, the closed sets $Y$ and $Z=\{x:|f(x)|\ge 1\}$ are disjoint. Choose by Urysohn's lemma a continuous $u:X\to [0,1]$ with $u{|}_Y=0$ and $u{|}_Z=1$, and defineThis function is continuous and agrees with $f$ on $Y$. On $Z$ it vanishes. Off $Z$ we have $|f|<1$ and $0\le 1-u\le 1$, so $|\tilde{f}|\le |f|<1$. Hence its range lies strictly inside $(-1,1)$ on all of $X$. The construction also covers empty $Y$ or empty $Z$ by a constant separating function where appropriate.