Are the elements of a set within a set also the elements of the latter? It is my understanding that an event is a subset of the set of all possible outcomes (sample space). If however the sample space consists of elements which are sets, can an event be defined as one the elements from these "inner" sets?
Ex. A coin is flipped twice, (S={(H,T),(T,H),(T,T),(H,H)} Is the event A=(H) valid for the sample space despite not being a subset of S?
 A: No.
In the first place, if $x\in y$ and $y\in z,$ then in most cases it is not true that $x\in z.$
For example, consider the set $\{\  \{1,2,3\},\  \{2,3,4\}\  \}.$ This set has only two members. If $1,2,3,4$ were members of it, then it would have at least four members.
In the second place, in the set $\{\,(H,T),(T,H),(T,T),(H,H)\,\},$ the pair $(H,T)$ is not a set with members $H$ and $T$; rather it is an ordered pair with components $H$ and $T.$
A: No, but you can get something close.  When you're looking a series of outcomes like that, the sample space is the product space of the sample space of each individual experiment.  Here, for example, your sample space is $\{H,T\}^2$, and if $\pi_i(a_1, \ldots, a_i, \ldots, a_n) := a_i$ is the $i$-th component function, you can arbitrarily define the event $E = \pi_1^{-1}(H)$ to get the event you're looking for.
More importantly, this set is well-behaved, meaning you can always find its probability (i.e. in terms of measure theory, it's measurable with respect to the product sigma algebra).
A: 
$\def\T{\mathcal T}\def\H{\mathcal H}$It is my understanding that an event is a subset of the set of all possible outcomes (sample space). If however the sample space consists of elements which are sets, can an event be defined as one the elements from these "inner" sets?
Ex. A coin is flipped twice, $S=\{(\H,\T),(\T,\H),(\T,\T),(\H,\H)\}$ Is the event $A=(\H)$ valid for the sample space despite not being a subset of S?

Well, no $(\H)$ is not an element of the sample space, nor is $\{(\H)\}$ a subset of it.   (Neither are $\H$ or $\{\H\}$ such, respectively.)
However, if by $A=(\H)$ you mean such as "the event of the first coin toss being heads", then this actually represents the event $\{(\H,\T), (\H,\H)\}$, which is indeed a subset of the sample space.

Let $A$ be a random variable mapping the sample space, $S$, to the result of the first toss of the coin, symbolically $A:S\mapsto \{\H,\T\}~,~\forall (x,y)\in S ~ [A(x,y):=x]$.
When we write $\mathsf P(A=x)$ we are actually referring to the event, $\{\omega\in S: A(\omega)=x\}$ .
It is convenient shorthand used because we really don't want to write the entire set builder notation every time.   It is also somewhat easier to read.
$$\begin{align*}\mathsf P(A=\H) &= \mathsf P\{\omega\in S: A(\omega)=\H\} \\ &= \mathsf P\{(\H,y)\in S:\forall y\in\{\H,\T\}\} \\ &= \mathsf P(A^{-1}(\H))  \\ &= \mathsf P_A(\H)\\&= \mathsf P\{(\H,\T), (\H,\H)\} \\&= \tfrac 12\end{align*}$$
A: No. Consider {{1}}, whose only element is {1}. 1 is an element in {1} but not in {{1}}.
