Ciaran said:
Ah I see, so how would I go about doing that?
Intuitively, equivalence classes are sets of equivalent elements, and elements are equivalent iff they belong to the same $S_i$. Therefore, equivalence classes are $S_i$ for $i=1,\dots,k$.
More precisely, you need to review the definition of equivalence classes of any given relation. We have $S=\bigcup_{i=1}^k S_i$. Let's suppose the definition says that the equivalence class of $x\in S$ with respect to $R$, denoted by $[x]$, is $\{y\in S\mid R(x,y)\}$ (the definition in your textbook or lecture notes may be slightly different, but essentially equivalent). Since $x\in S$, there is some $1\le j\le k$ such that $x\in S_j$. Since $S_i$ form a partition, they are disjoint, so $x\notin S_i$ for any $i\ne j$. Then
\begin{align}
R(x,y)&\iff\exists i\;(x\in S_i\text{ and }y\in S_i)&&\text{by definition of }R\\
&\iff y\in S_j&&\text{because }x\text{ is in }S_j\text{ and only }S_j.
\end{align}
Therefore,
\[
[x]=\{y\in S\mid R(x,y)\}=\{y\in S\mid y\in S_j\}=S_j
\]
Since $x$ was arbitrary and all equivalence classes have the form $[x]$ for some $x\in S$, every equivalence class is one of $S_i$. Conversely, by definition of partition, each $S_j$ is nonempty, so there exists an $x\in S_j$, and then, as shown above, $[x]=S_j$. So all $S_i$ are equivalence classes.
Now suppose there are two relations $R_1$ and $R_2$. Let's denote the equivalence classes of $x$ with respect to $R_1$ and $R_2$ by $[x]_1$ and $[x]_2$, respectively. We are given that $S_1,\dots,S_k$ are $[x]_1$ for various $x$, and similarly for $[x]_2$. Fix some $x\in S$ and suppose $j$ is the unique index such that $x\in S_j$. Then $[x]_1=[x]_2=S_j$ (Why?). Therefore,
\[
R_1(x,y)\iff y\in [x]_1\iff y\in [x]_2\iff R_2(x,y),
\]
so $R_1$ coincides with $R_2$.
