I am trying to understand why the following statement is true:
If $S$ and $S'$ are subsets of $\mathbb{K}[X_1,...,X_n]$ such that $S\subseteq S'$, then $\mathcal{V}(S') \subseteq \mathcal{V}(S)$.
Can someone provide an example of two polynomial sets with their varieties, such that I might understand why this is true.
What you need to realize is that intuitively, the vanishing sets $V(S)$ are essentially the embeddings of your polynomials into $n$-dimensional space.
So, for instance, let's take $S=\{Y-2X\}\subseteq \mathbb{R}[X,Y]$. Then
$$V(S)=\{(x,y)\in\mathbb{R}^2\mid y-2x=0\}=\{(x,y)\in\mathbb{R}^2\mid y=2x\}$$
and you see that $V(Y-2X)$ just represents the line $y=2x$ in the plane.
What happens if we have more than one polynomial? Then it becomes where your curves intersect in the plane.
So, let $S'=\{Y-2X,Y-4\}$. Then
$$V(S')=\{(x,y)\in\mathbb{R}^2\mid y-2x=y-4=0\}=\{(2,4)\}$$
where we get the point $(2,4)$ from solving the equation $y-2x=y-4=0$. This is representing the fact that $(2,4)$ is the only point of intersection of the two lines $y=2x$ and $y=4$.
Of course, this point is a subset of the set of solutions to $y=2x$, so in particular we see $V(S')\subset V(S)$.