$\underline{\textbf{Given}}$
$A$, $B$ are $\underline{\text{finite}}$ sets, such that $A \subseteq B$. We need to find $\sum_{C: A \subseteq C \subseteq B} (-1)^{|C-A|}$
$\underline{\textbf{Venn Diagram}}$
Let $|B| = n$ and $|A|=k$, then $|C-A|=m$
$\underline{\textbf{Observations}}$
For $|C-A|=m$ we only need to consider the count of such $m$ values and how many distinct types of $m$ values can be formed. But why we doing this? $m$ value will decide if $(-1)^m$ will be $1$ or $-1$, and count of specific $m$ value will decide the count of $1$ or $-1$. This can significantly reduce the above problem to simpler version, easing the further process of getting final solution.
According to above figure $k \le |C| \le n$ same as $k \le \underbrace{\overbrace{|Y_1|}^{m}+\overbrace{|A|}^{k}}_{\substack{C = Y_1+A \\ Y_1 \cap A = \phi}} \le n$, this implies $\underbrace{0 \le m \le n-k}_{\text{distinct m values}}$. And for specific $m$ value, count is $^{n-k}C_m$.
$\underline{\textbf{Calculations}}$
$$\begin{align}
\sum_{C: A \subseteq C \subseteq B} (-1)^{|C-A|} &= \sum_{m=0}^{n-k} \ ^{n-k}C_m (-1)^m \\
&= (1+(-1))^{n-k} \tag{Binomial Expansion} \\
&= 0^{n-k}
\end{align}$$
$\underline{\textbf{Final Solution}}$
$0^{n-k} = \begin{cases}1 & n = k \implies A = B \\ 0 & n \neq k \implies \text{otherwise}\end{cases}$
$\textbf{Option (D) is correct}$
$\underline{\textbf{Reference for }0^0=1}$
Interesting read "What is 0^0"