Yes. Choose a reduced expression $red(w)$ for $w$. Now delete from $red(w)$ the rightmost letter $s_i$ that appears more than once in $red(w)$. This will yield a reduced expression for an element $v\in W$ with the same support as $ w$ and length one less than $w$, with $v$ less than $w$ in Bruhat order. Repeat until you have decreased the length to $r+1$.

Added later: To prove that my claim holds for an arbitrary reduced expression, rather than making a special choice, one can use the exchange axiom for Coxeter groups as follows. Suppose the new expression obtained by deleting $s_i$ is not reduced, then consider the leftmost letter $s_{i_t}$ to the right of $s_i$ such that $s_{i_1} \cdots \hat{s_i} \cdots s_{i_t}$ is nonreduced. Now consider the rightmost $r$ so that $s_{i_r}\cdots \hat{s_i} \cdots s_{i_t}$ is nonreduced. Then $s_{i_r}\cdots \hat{s_i} \cdots s_{i_{t-1}}$ and $s_{i_{r+1}}\cdots \hat{s_i} \cdots s_{i_t}$ will be reduced expressions for the same Coxeter group element. This implies that the former expression must contain the letter $s_{i_t}$, a contradiction to $s_{i_t}$ only appearing once in $red(w)$.

Yes. Choose a reduced expression $red(w)$ for $w$. Now delete from $red(w)$ the rightmost letter $s_i$ that appears more than once in $red(w)$. This will yield a reduced expression for an element $v\in W$ with the same support as $ w$ and length one less than $w$, with $v$ less than $w$ in Bruhat order. Repeat until you have decreased the length to $r+1$.

Added later: To prove that my claim holds for an arbitrary reduced expression, rather than making a special choice, one can use the exchange axiom for Coxeter groups as follows. Suppose the new expression obtained by deleting $s_i$ is not reduced, then consider the leftmost letter $s_{i_t}$ to the right of $s_i$ such that $s_{i_1} \cdots \hat{s_i} \cdots s_{i_t}$ is nonreduced. Now consider the rightmost $k$ so that $s_{i_k}\cdots \hat{s_i} \cdots s_{i_t}$ is nonreduced. Then $s_{i_k}\cdots \hat{s_i} \cdots s_{i_{t-1}}$ and $s_{i_{k+1}}\cdots \hat{s_i} \cdots s_{i_t}$ will be reduced expressions for the same Coxeter group element. This implies that the former expression must contain the letter $s_{i_t}$, a contradiction to $s_{i_t}$ only appearing once in $red(w)$.

Yes. Choose a reduced expression $red(w)$ for $w$. Now delete from $red(w)$ the rightmost letter $s_i$ that appears more than once in $red(w)$. This will yield a reduced expression for an element $v\in W$ with the same support as $ w$ and length one less than $w$, with $v$ less than $w$ in Bruhat order. Repeat until you have decreased the length to $r+1$.

Yes. Choose a reduced expression $red(w)$ for $w$. Now delete from $red(w)$ the rightmost letter $s_i$ that appears more than once in $red(w)$. This will yield a reduced expression for an element $v\in W$ with the same support as $ w$ and length one less than $w$, with $v$ less than $w$ in Bruhat order. Repeat until you have decreased the length to $r+1$.

Added later: To prove that my claim holds for an arbitrary reduced expression, rather than making a special choice, one can use the exchange axiom for Coxeter groups as follows. Suppose the new expression obtained by deleting $s_i$ is not reduced, then consider the leftmost letter $s_{i_t}$ to the right of $s_i$ such that $s_{i_1} \cdots \hat{s_i} \cdots s_{i_t}$ is nonreduced. Now consider the rightmost $r$ so that $s_{i_r}\cdots \hat{s_i} \cdots s_{i_t}$ is nonreduced. Then $s_{i_r}\cdots \hat{s_i} \cdots s_{i_{t-1}}$ and $s_{i_{r+1}}\cdots \hat{s_i} \cdots s_{i_t}$ will be reduced expressions for the same Coxeter group element. This implies that the former expression must contain the letter $s_{i_t}$, a contradiction to $s_{i_t}$ only appearing once in $red(w)$.