Question 1. Suppose $H$ is a nonempty subset of a finite (multiplicative) group $G$, such that whenever $a,b \in H$, the element $a \cdot b$ is also in $H$. We want to show that $H$ is a subgroup of $G$. That is, we want to show that $(H, \cdot)$ is a group. We verify the group axioms for $H$.
Closure under the $\cdot$ operation. If $a,b$ are elements of $H$, their product $a \cdot b$ is guaranteed to lie in $H$. So $H$ is indeed closed under $\cdot$. So there's nothing to prove.
Associativity. Suppose $a,b,c$ are elements of $H$, and hence of the group $G$. Then since the $\cdot$ is associative (remember that $G$ is a group), we have $a \cdot (b \cdot c) = (a \cdot b) \cdot c$. Thus the subgroup $H$ inherits the associativity of $\cdot$ from $G$. [Still nothing to prove! Don't we wish all proofs are as simple? :-)]
Existence of identity element $e$ in $H$. In this case, there is something to prove, but we will come to this after looking at inverses.
Existence of inverses. Again, there's something to prove. This was the point that was explicitly shown in great detail in the lecture (through the pigeonhole principle argument).
Coming back to item (3.), note that we still have not established that the identity element $e$ is in $H$. Let $a$ be an arbitrary element of $H$ (here we need the hypothesis that $H$ is nonempty). From item (4.), the inverse $a^{-1}$ is in $H$. Since $H$ is closed w.r.t. $\cdot$, it follows that $e = a \cdot a^{-1} \in H$ as well. $\Box$
Question 2. Suppose $a$ is in $H$. Then $a^2 = a \cdot a$ is also in $H$ by closure property. Then $a^3 = a \cdot a^2$ is also in $H$. Similarly, $a^4 = a \cdot a^3$ is also in $H$, and so on. More generally, we can use mathematical induction to prove the proposition that $a^n$ is in $H$ for each natural number $n$. (For the induction step, we need to show that if $a^{n-1} \in H$, then $a^n \in H$ as well. But this is true because $a^n = a \cdot a^{n-1}$ and $H$ is closed under the $\cdot$ operation.)
Question 3. Consider the sequence $\langle a, a^2, a^3, \ldots, a^n, \ldots \rangle$ of infinite length. Since the group $G$ is finite, there are only a finite number of distinct terms in that sequence. Therefore, some two terms, say the $i^{th}$ and $j^{th}$ terms, are equal. That is, $a^i = a^j$.
Question 4. That conclusion is false. What is true is that if $a$ is in $H$, then $\langle a \rangle \subseteq H$.