The Partition Problem: Reducing to prove NP-Completeness
Partition is equivalent to a special case of the subset-sum problem: $S$ can be partitioned into two sets $A$ and $B$ such that $\sum_{x \in A} x = \sum_{x \in B} x$ if and only if $\exists S' \subseteq S$ such that $\sum_{x\in S'} x = \frac{1}{2}\sum_{x \in S}x$. 

Proof of $\implies$: Given $A$ and $B$ that partition $S$ and are such that $\sum_{x \in A} x = \sum_{x \in B}$ you must have $\sum_{x \in S}x =\sum_{x \in A} x + \sum_{x \in B} x = 2 \sum_{x \in A} x \implies \sum_{x \in A} x = \frac{1}{2}\sum_{x \in S}x$. The claim follows by picking $S' = A$.

Proof of $\Longleftarrow$: Given $S' \subseteq S$  such that $\sum_{x\in S'} = \frac{1}{2}\sum_{x \in S}x$ you must have $\sum_{x\in S \setminus S'} x = \sum_{x \in S}x - \sum_{x \in S'}x = \frac{1}{2}\sum_{x \in S}x$. The claim follows by picking $A=S'$ and $B = S \setminus S'$.



To show that this special case of subset sum (i.e. partition) is in $\mathsf{NP}$ notice that a subset $S' \subset S$ such that $\sum_{x \in S'} x = \frac{1}{2}\sum_{x \in S}$ is a yes-certificate.

To see that this special case of subset sum is still $\mathsf{NP}$-hard consider an instance $\langle S, T \rangle$ of subset sum where $S$ is a (multi-)set and $T$ is the target value. Let $M = \sum_{x \in S} x$.

Create a new instance subset sum $\langle \overline{S}, \overline{T} \rangle$ where $\overline{S} = S \cup \{ M-2T \}$ and $\overline{T}=M-T$.

If there is a solution $S' \subseteq \overline{S}$ to $\langle \overline{S}, \overline{T} \rangle$ then $\sum_{x \in S'} x = \sum_{x \in \overline{S} \setminus S'} x= M-T$ (since the sum of the elements in $\overline{S}$ is $2(M-T)$). At least one of $S'$ and $\overline{S} \setminus S'$ contains element $M-2T$, call this set $S^*$. Then the $S^* \setminus \{M-2T\}$ is a subset of $S$ and the sum of its elements is $(M-T)-(M-2T)=T$, showing that the original instance $\langle S, T \rangle$ is a yes instance.

On the other hand, if there is a solution $S' \subseteq S$ to $\langle S, T \rangle$, then the set $S' \cup \{ M-2T \}$ is a subset of $\overline{S}$ and the sum of it's elements is $T + (M - 2T) = M-T = \overline{T}$, showing that $\langle \overline{S}, \overline{T} \rangle$ is a yes instance. This concludes the proof.