The Knapsack problem is NP, and any problem in NP can be reduced to an NP complete problem (Cook's Theorem). So to show that the knapsack problem is NP complete it is sufficient to show that an NP-complete problem is reducible to the Knapsack problem. We want to use the exact cover problem to show this. A lot of such reductions can be found in the paper of Karp[1], including the reduction that is described here.
My notation conforms mostly to the notation in this paper.
Assume that $\{S_j,j=1,\ldots,p\}$ is a family of sets that covers the set
$U=\{u_i,i=1,\ldots,t\}$, so
$$\cup_{j=1}^{m} S_j=U$$
We want to check if there is a subfamily $\{S_{j_h},h=1,\ldots,q\}$ of $\{S_j,j=1,\ldots,p\}$ that covers $U$ and where the members of these family are pairwise disjoint. Let $d$ be an arbitrary positive integer. We can assign the number
$$\; \epsilon_{i,j}= \begin{cases}
0,& u_i \not \in S_j \\
1, & u_i \in S_j
\end{cases} \\
a_j=\sum_{j=1}^m \epsilon_{i,j} d^{i-1} \tag{1}$$
to a subset $S_j$.
So $a_j$ is a number that can be written only with digits $0$ and $1$ in a number system with base $d$. It has the digit $1$ on position $i-1$ if and only if $u_i \in S_j$. If $a_{j_1}$ and $a_{j_2}$ are to values assigned to two sets of an exact cover, then there is no position in the representation as an $d$ base number, where both numbers have the digit $1$. The sum $b$ of all the numbers assigned to an exact cover is
$$b=\sum_{i=1}^{t}d^{i-1} \tag{2}$$
if $\{S_j,j=1,\ldots,p\}$ has an exact cover $\{S_{j_h},h=1,\ldots,q\}$ then $x=(x_1,\cdots,x_p)$ with
$$ x_j=
\begin{cases}
1 & \text{if}\; j=j_h, \text{for an } h \in \{1,...,q\} \\
0 & \text{if}\; j\ne j_h, \text{for all } h \in \{1,...,q\} \\
\end{cases}$$
is one solution of the
of the 0-1-Knaspack problem
$$\sum_{i=1}^{p} a_i x_i=b$$
where $a_j$ and $b$ are defined in $(1)$ and $(2)$.
If we select the base $d=p+1$ (or larger) fo all the $2^p$ 0-1-sums no carry will occurr when adding them. So when ome of them sum up to be, there summands con't have 1s on the same place.
[1] Richard M. Karp: Reducibility Among Combinatorial Problems
In: R. E. Miller and J. W. Thatcher (Hrsg.): Complexity of Computer Computations.
Plenum Press, New York, 1972,
p. 85–103
Best Answer
Partition can be reduced to this problem. Set $S$ can be partitioned iff set of rectangles $\{1 \times (4\cdot x)| x \in S\}$ can be packed into rectangle $2 \times (\sum_{x \in S} 2x)$: rectangles from upper row corresponds to one subset, from bottom to other, and multiplication by $4$ ensures all rectangles are placed horizontally.