I agree with the above answer, but to turn it into a rigorous proof one could argue in a slightly different fashion:

Let $l$ be the length of $w$ and let $Red(w)$ be the set of $all$ reduced expressions for $w$. Given $r=(i_1,\ldots, i_l)\in Red(w)$ denote by $k(r)$ the smallest $k\le l$ such that $i_k$ appeas in $r$ more than once. Let $t=(j_1,\ldots, j_l)\in Red(w)$ be such that $k(t)\ge k(r)$ for all $r\in Red(w)$. Write $w=s_{j_1}\cdots s_{j_l}$ and let $w'$ be the element of $W$ obtained from $w$ by deleting $s_{j_k}$ from the above expression where $k=k(t)$. By the choice of $k$ any reduced expression for $w'$ will have full support. Repeat this step until the length decreases to $r+1$.

I agree with the answer below, but to turn it into a rigorous proof one could argue in a slightly different fashion:

Let $l$ be the length of $w$ and let $Red(w)$ be the set of $all$ reduced expressions for $w$. Given ${\bf r}=(i_1,\ldots, i_l)\in Red(w)$ denote by $k({\bf r})$ the smallest $k\le l$ such that $i_k$ appeas in $\bf r$ more than once. Let ${\bf t}=(j_1,\ldots, j_l)\in Red(w)$ be such that $k({\bf t})\ge k({\bf r})$ for all ${\bf r}\in Red(w)$. Write $w=s_{j_1}\cdots s_{j_l}$ and let $w'$ be the element of $W$ obtained from $w$ by deleting $s_{j_k}$ from the above expression where $k=k({\bf t})$. By the choice of $k$ any reduced expression for $w'$ will have full support. Repeat this step until the length decreases to $r+1$.