Is $X\to\{p\}$ a proper morphism precisely when $X$ is a projective variety? For me, the notion of a proper morphism is a map $\varphi\colon X\to Y$ such that $\varphi$ factors as $X\xrightarrow{i}Y\times\mathbb{P}^m\xrightarrow{\pi} Y$, where $\mathbb{P}^m$ is some projective space over an algebraically closed field, $i$ is a closed embedding, and $\pi$ the projection on to $Y$.
I think that a morphism $\varphi\colon X\to\{p\}$ to a point is proper iff $X$ is a projective variety. If $X$ is a projective variety, then it is closed in some $\mathbb{P}^m$. So we can embed $X\hookrightarrow\mathbb{P}^m\hookrightarrow\{p\}\times\mathbb{P}^m$, which is a closed embedding since the first map is inclusion, and the second map is just $x\mapsto (p,x)$. Composing with the projection onto $p$ gives back $\varphi$. 
On the other hand, if $\varphi\colon X\to\{p\}$ is a proper morphism, then there exists a closed embedding $X\to\{p\}\times\mathbb{P}^m$, so $X$ is (isomorphic) to a closed subvariety of $\{p\}\times\mathbb{P}^m$, and hence a closed subvariety of $\mathbb{P}^m$. And so I conclude $X$ is a projective variety?
Does this look reasonable? Thanks.

No: a variety such that  $X\to \{p\}$ is proper  is said to be complete.
A complete variety of dimension $1$ is indeeed projective  but for any $d\geq 2$ there exist complete and non projective varieties of dimension $d$.
In dimension $2$ a complete smooth variety must be projective (Kodaira) , but  there exists a  smooth complete variety of dimension $3$ which is not projective.
This was proved by Hironaka and you can see such a variety [for some reason it is blue ...] on the cover of Shafarevich's Basic Algebraic Geometry, vol.2.
The details of Hironaka's construction are given  in that volume, starting page 75. 
