For your three numbered statements, $(1)$ is true (easy proof), but $(2)$ and $(3)$ are false. A counter example for $(2)$ is obtained by taking $X = (1,2)$ and $Y = (3,4)$. Then $XY = (1,2)(3,4)$ is an involution as well. A counterexample for $(3)$ is obtained from this example as well; the involution $X = (1,2)(3,4)$ can be factored as $YZ$ where $Y = (1,2)$ and $Z = (3,4)$.
As for the statement in question, here's a quick proof sketch:
(1) By using the disjoint cycle decomposition, you can reduce to proving that the cycle $(1,2,3,\dots,n)$ can be written as a product of two involutions in $S_n$.
(2) To handle that case, draw $n$ vertices in the plane (labelled $1,2,\dots,n$) and connect the $n$ vertices by drawing $n-1$ edges. This will make a unique (up to choice of direction to travel) path in your graph. Label the edges $1,2,\dots,n-1$ in the order of the path. For each edge, put the two vertices connected by that edge into a two-cycle. Then form $\pi_1$, the product of the two-cycles formed in this way from odd-numbered edges, and $\pi_2$, the product of the two-cycles formed in this way from even-numbered edges. Then the product $\pi_2 \pi_1$ is an $n$-cycle $\tau$. This needs to be checked; in fact, if you number the vertices in the order of the path, then $\tau = (1,3,5,\dots, 6,4,2)$. Conjugate the relation $\tau = \pi_2 \pi_1$ to get that $(1,2,\dots,n)$ is a product of two involutions.
The proof is indeed correct. There is a nice intuition behind this:
You can interpret a permutation as a directed graph: The nodes are the numbers $1,\dots,n$ and there is an arrow from $a$ to $b$ if the permutation $\pi$ maps $\pi(a)=b$. This is a functional graph and as the number of nodes is finite and the map is one-to-one, every node has exactly one in-edge and exactly one out-edge. It is clear that this graph must be a set of disjoint cycles (hence the name).
Of course this is only a picture to the proof, to show where it comes from. The proof is correct because we can easily check that $S_n$ indeed acts on $X$ and orbits are cycles.
EDIT: Someone requested a real picture, not only a sketch of the proof. So I add one showing the corresponding graph of the permutation $\left(\begin{eqnarray} 1&2&3&4&5&6&7&8&9&10 \\ 8&6&5&9&3&10&4&7&1&2 \end{eqnarray}\right)$.
Best Answer
It might help to consider an example. Suppose we have the following disjoint cycle decomposition of a permutation: $$ \sigma = (1 \ 7\ 4\ 10)(2\ 9\ 8)(3\ 5\ 6). $$ To begin, separately decompose each cycle: $$ (1 \ 7\ 4\ 10) = \tau_{1,1}\tau_{1,2}, \quad (2\ 9\ 8) = \tau_{2,1}\tau_{2,2} \quad (3\ 5\ 6) = \tau_{3,1} \tau_{3,2}. $$ Note that $\tau_{1,1},\tau_{1,2}$ are permutations that only affect the elements $1,4,7,10$. Similarly, $\tau_{2,1},\tau_{2,2}$ only affect $2,9,8$. In other words, for any $i \neq j$, the elements $\tau_{i,p}, \tau_{j,q}$ are disjoint permutations, which means that $\tau_{ip}\tau_{jq} = \tau_{jq}\tau_{ip}$.
With that established, we can use this commutativity property to "move" the transpositions $\tau_{i,1}$ to the left. That is, we can write $$ \sigma = \tau_{11}(\tau_{12}\color{red}{\tau_{21}})\tau_{22}\tau_{31}\tau_{32}\\ = \tau_{11} (\color{red}{\tau_{21}}\tau_{12}) \tau_{22}\color{red}{\tau_{31}} \tau_{32}\\ = \tau_{11}\tau_{21}\color{red}{\tau_{31}}\tau_{12}\tau_{22}\tau_{32}\\ = (\tau_{11}\tau_{21}\tau_{31})(\tau_{12}\tau_{22}\tau_{32}). $$ Now, we see that $\sigma$ is a product of the involutions $\tau_{11}\tau_{21}\tau_{31}$ and $\tau_{12}\tau_{22}\tau_{32}$.