I have already read this post, which answers the problem by first delving into the elliptization conjecture, which seems to me to be hiding details, because I don't understand how the elliptization conjecture is a special case of the Geometrization conjecture (now Theorem I suppose). Let us consider the following statement of Thurston's theorem:
Theorem 1: Let $M$ be a closed, orientable, prime $3$-manifold. Then there exists an embedding of a disjoint union of incompressible $2$-tori and Klein bottles in $M$ such that every component of their complement admits a locally homogeneous Riemannian metric of finite volume, isometric to one of the $8$ model geometries.
We also have the following lemma
Lemma 1: If $M$ is a Riemannian Manifold, with universal cover $X$, which admits a complete locally homogeneous Riemannian metric then $M$ is isometric to $X/\Gamma$, where $\Gamma$ is the deck group of $X$.
Now my attempt to understand how the Geometrization theorem implies the Poincaré conjecture is as follows:
Let us now consider a compact, simply connected $3$ manifold $M$. As it has a trivial fundamental group, the decomposition referred to in Theorem 1 must be trivial (because Tori and Klein bottles have non trivial fundamental groups, so there can be no such incompressible embedded surfaces.) Thus $M$ must have a complete locally homogeneous metric. Lemma 1 thus implies that $M\simeq X/\Gamma$, where $X$ is its universal cover, and $\Gamma$ its deck group. The deck group is isomorphic to $N(H_0)/H_0$, where $N$ is the normalizer, and $H_0$ is the pushforward of the fundamental group of $X$ under the covering map. As the fundamental group is trivial this then implies that the deck group is trivial, hence $M$ is isometric to its universal cover $X$.
Now I'm hoping that a simply connected $3$ manifold being isometric to its universal cover means it has to be diffeomorphic to $S^3$, but so far have found no evidence to support my hope. Am I on the right track? Or do I need to consider something fundamentally different?
As is apparent from this post I am a novice in the areas of differential geometry and algebraic topology, so would appreciate it if answers used as simple tools as possible. I am also aware of the Kesner decomposition theorem, but I am sure understand none of its subtleties.
New Attempt:
Following the informative comment of Mike Miller and answer from Eric Towers I present their argument with some of the gaps filled in, just in-case any novices like me stumble across this post, as I did not immediately understand how their argument worked.
Let us consider a compact, simply connected $3$ manifold $M$. As it has a trivial fundamental group, the decomposition referred to in Theorem 1 must be trivial (because Tori and Klein bottles have non trivial fundamental groups, so there can be no such incompressible embedded surfaces.) Thus $M$ must have a complete locally homogeneous metric. This implies that $M\simeq X/\Gamma$, where $X$ is one of the $8$ Thurston geometries, and $\Gamma$ a group of isometries of $X$ acting transitively on $X$. Thus $X/\Gamma$ is also simply connected. The quotient map $\pi:X\to X/\Gamma$ is a regular covering map, because the action of $\Gamma$ is properly discontinuous [1]. As each of the model geometries is path connected, so too is $X/\Gamma$, which means $X$ is homeomorphic to $X/\Gamma$ due to the latter being simply connected. Hence $M$ is homeomorphic to $X$, meaning $X$ is compact, but $S^3$ is the only compact model geometry.
[1] Munkres, James R., Topology, Pearson Education (2000).
I recommend reading the first two sections of J.W. Morgan's "Recent Progress on the Poincare Conjecture and the Classification of 3-Manifolds". These pages cleared up your question for me. The discussion there is in the same concept space as Mike Miller's comment.