Topology of a cone of $\mathbb R\mathbb P^2$.
It's not a manifold either with or without boundary. For the proof, look at the cone point c. If U is a small open set around c, then $U-c$ is homeomorphic to $\mathbb{R}P^2 \times (0,1]$. However, this is not homeomorphic to either $\mathbb{R}^3-pt$ nor a half space minus a point.
In fact, it's contractible (the cone over anything is contracitble). Thus, it's simply connected.
I'm not sure, but it can definitely be topological embedded in R^5. To see this, let $f$ be an embedding of $\mathbb{R}P^2$ into $\mathbb{R}^4$ (which exists by Whitney's embedding theorem). Define $G:Z\rightarrow\mathbb{R}^5$ by $G(p, t) = (tf(p),t)$. This will be a topological embedding.
The minimal possible answer is 4 since that's the smallest in which $\mathbb{R}P^2$ can be embedded into.
First consider just the hemisphere $X'\subseteq X$, and define $Z'= X'/E$. This is the real projective plane, in one of its many incarnations. You may have seen $\mathbb{RP}^2$ presented as the space of lines through the origin in $\mathbb{R}^3$. To get from that presentation to this one, note that each such line intersects $S^2 \subseteq \mathbb{R}^3$ in exactly two points, and for those lines where one point is above the xy-plane and the other is below, we've just chosen the $z>0$ representative. This is a manifold without boundary, and it is not simply-connected. Indeed, its universal cover is $S^2$, which is a two-fold cover. Hence its fundamental group has two elements.
Now, note that in $Z$, we're literally just taking the cone over $Z'$ -- those rays from the origin that end at points in $X'$ that were identified in $Z'$ are themselves identified. It is a non-trivial (?) fact that $\mathbb{RP}^2$ is not the boundary of any 3-manifold. I'm pretty sure that if the cone over a space is a manifold, then the boundary is the manifold itself, although I can't instantly tell you why this should be true. (In fact, $\mathbb{RP}^2$ generates the unoriented cobordism ring in dimension 3.) Note that the cone over any space is simply-connected, since it is contractible (to the cone point). I don't know the answer to 4, though I'd suspect it's either 4 or 5, since 4 is the minimum dimension you need in order to embed $\mathbb{RP}^2$. In general, the question of minimum dimension for embedding a particular manifold is not so easy, beyond the a priori bounds given by the Whitney embedding theorem. And this isn't even a manifold...
This should be a comment somehwere, but got to long.
Will, the cone on $P^2_\mathbb R$ is not a manifold. Consider the long exact sequence for integral reduced homology of the pair $(C,C')$ where $C=C(P^2_\mathbb R)$ is the cone over $P^2_\mathbb R$ and $C'=C\setminus\{a\}$ is the complement of the apex of the cone. Since $C$ is contractible and $C'$ deformation-retracts onto $P^2_\mathbb R$, you get isomorphisms $H^\sharp_2(C,C')\cong H^\sharp_{1}(P^2_\mathbb R)\cong\mathbb Z/2\mathbb Z$.
If follows that $C$ is not a manifold: in a manifold $M$, for every point $p\in M$ we have that the integral reduced homology $H^\#_\bullet(M,M\setminus\{p\})$ is that of a sphere.
(Generalizing this reasoning, you get a rather strong condition on a manifold for its cone to be also a manifold. I'm sure the topologists among our fellow M.SEers know of a precise characterization.)
In the same way, we see that $C$ is not a manifold with boundary, because in such a space $H^\#_\bullet(M,M\setminus\{p\})$ is, for every $p$, either identically zero or that of a sphere.