In the case of very simple expressions like this, an alternative approach is to use "truth tables". Essentially, one considers, for an element $x$, all possible combinations of whether it is or it not in any given set, and computes whether it is in the set determined by the complex expression. Here, with three sets, we need to consider $8$ possibilities.
Use $0$ to denote that the element is not in the set, and $1$ to denote that it is. Remembering that $x\in X-Y$ if and only if $x\in X$ and $x\notin Y$, we have: $\begin{array}{|c|c|c||c|c|} \hline A&B&C&A-B&(A-B)-C\\ \hline 0 & 0 & 0& 0 & 0\\ \hline 0 & 0 & 1 & 0 & 0\\ \hline 0 & 1 & 0 & 0 & 0\\ \hline 0 & 1 & 1 & 0 & 0\\ \hline 1 & 0 &0 & 1 & 1\\ \hline 1 & 0 & 1 & 1 & 0\\ \hline 1 & 1 & 0 & 0 & 0\\ \hline 1 & 1 & 1 &0 & 0\\ \hline \end{array}$
For $A-(B\cup C)$, we have: $\begin{array}{|c|c|c||c|c|} \hline A&B&C&B\cup C&A-(B\cup C)\\ \hline 0 & 0 & 0& 0 & 0\\ \hline 0 & 0 & 1 & 1 & 0\\ \hline 0 & 1 & 0 & 1 & 0\\ \hline 0 & 1 & 1 & 1 & 0\\ \hline 1 & 0 &0 & 0 & 1\\ \hline 1 & 0 & 1 & 1 & 0\\ \hline 1 & 1 & 0 & 1 & 0\\ \hline 1 & 1 & 1 &1 & 0\\ \hline \end{array}$
Now look at the final column: since the columns for $A-(B\cup C)$ and for $(A-B)-C$ are identical, it follows that the sets are the same.
Now, for large number of "sets", this technique becomes unwieldly: the table will have $2^n$ rows when the equation involves $n$ different sets. But for small numbers of sets (two or three, usually), it provides a simple, mechanical way of testing for equality. For example, the truth-table method is much simpler to prove the associativity of the symmetric difference than element-wise methods.