 Solution

For each point which belongs to at least one $U_t$, its membership-index set is a nonempty subset of $[0,1]$, so has an infimum in $[0,1]$. Assigning the value one to the other points therefore defines $f:X\to [0,1]$ everywhere.

For $0<r\le 1$,An index $t<r$ gives $f(x)\le t<r$. Conversely if the infimum is below $r$, some membership index must be below $r$. Thus this strict sublevel set is open.

For $0\le r<1$ we claimIf $x\notin \overline{U_t}$ and $t>r$, then $x$ cannot belong to any $U_s$ with $s\le t$, since $U_s\subseteq U_t$ for $s<t$. Hence $f(x)\ge t>r$. Conversely, if $f(x)>r$, density of $D$ supplies $r<t<u<f(x)$ with $t,u\in D$. Since $x\notin U_u$ and $\overline{U_t}\subseteq U_u$, one has $x\notin \overline{U_t}$. This proves the identity, and its right-hand side is open.

The remaining sublevel and superlevel sets, for thresholds outside these ranges, are either empty or all of $X$. Preimages of open rays are consequently open, and these rays generate the usual real topology. Therefore $f$ is continuous. The closure inclusion is what supplies openness of the strict superlevel sets; mere nesting would not suffice.