I have been able to prove to myself that the boundary of a 3-dimensional manifold is indeed a compact set. I am stuck however proving that it is a 2 dimensional manifold. Specifically why the following each point in the boundary is contained in some $V$ and why this $V$ is open.
Let $X$ be a compact manifold with boundary.
I understand that there is an open set $U \subseteq Bdy(X)$ where $h$ is a homeomorphism such that $h:U \to \{(x_1,x_2,x_3)\in\mathbb{R}^3\,:x_1^2+x_2^2+x_3^2<1\, , x_1\geq 0\}$ given by $h(x)=(0,0,0)$.
Let $V=h^{-1}[\{(x_1,x_2,x_3)\in\mathbb{R}^3:x_1^2+x_2^2+x_3^2<1\, , x_1=0\}]$. Let $x \in Bdy(X)$. Then $x \in V$ because if we fix an $x$ on the open ball, we can find a great circle that contains $x$. This is open because the complement $Bdy(X)\setminus V$ is closed.
Am I on the right track?
You're on the right track. I think your confusion is in your understanding that "there is an open set $U\subseteq Bdy(X)$ where $h$ is a homeomorphism ..." --- it's not clear what the quantifiers are here, nor how $h$ is defined (is it to be constant?).
Here's a hint to get you back on course. Recall from the definition of "manifold with boundary" that every $x\in X$ is contained in a chart neighborhood homeomorphic to a three-ball or in a chart neighborhood homeomorphic to a half-ball.
First characterize $\partial X$ as those points which are only contained in atlas neighborhoods of the second type. Then prove that $x\in\partial X$ implies the image of $x$ under a chart neighborhood is contained in $(\mathbb{R}^2\times\{0\})\cap B_0(1)$. Why are you now done?