Why is a Zero Vector Space a Vector Space? The first example in my Linear Algebra textbook states this is so by stating the example satisfies axiom $4$, which says there is zero vector, there is an object $u$ in which $0 + u = u + 0 = u$, where $0$ is the zero vector and $u$ is an object or a set of real number.
Is there not already a violation in axiom 1 where the zero space is not closed under addition because the vector space is missing another object v?
 A: The zero set is closed under addition: $0+0=0$. It is closed under scalar multiplication: $\lambda\cdot 0=0$. It is non-empty. Hence it is a vector space, and in fact the only one of dimension $0$ (up to isomorphism and once the field is fixed).
A: Axiom 1, as you report it, is

if $u$ and $v$ are objects in $V$, then $u + v$ is in $V$

I wouldn't classify this among the axioms: it's just the requirement that an operation is defined on $V$. However, a correct reading of the statement is

for every $u\in V$ and every $v\in V$, $u+v\in V$

Nothing whatsoever prevents $u$ and $v$ from denoting the same object (or, better, from being substituted with the same object in $V$); are you going to doubt that $2+2$ can be performed in the natural numbers?
An operation on $V$ is just a map
$$
V\times V\to V
$$
and as such it must be defined also on pairs of the form $(x,x)$. Is there an operation
$$
\{0\}\times\{0\}\to\{0\}
$$
that makes $\{0\}$ into an abelian group? Surely so! There's only one map! Can you define a map $F\times\{0\}\to\{0\}$? Yes, of course, there's only one.
Is $\{0\}$ a vector space under these operations? Yes, indeed. Can you falsify the associative property or any other of the requested properties?
So, yes, any one element set is a vector space (on whatever field you fix), because there's only one way to define the operations on it.
A: The zero set satisfies the two properties of a subspace.
i) Closed under addition. $0+0=0$
ii) Closed under scalar multiplication $\lambda(0)=0$. 
My previous edit was naively wrong.
