How to prove a subspace is non empty? To prove that a subspace W is non empty we usually prove that the zero vector exists in the subspace. But then is it necessary to prove the existence of zero vector. Can't we  prove the existence of any vector instead?  
Can someone please explain with an example where we can prove that W is a subspace by taking the existence of any random vector?
 A: Well, in general if you want to prove that a set $S$ is not empty, then you just have to prove that it contains an element. This element can be the $0$ element or any other (this don't matter). 
Now, suppose that $V$ is a $\Bbb F$ vector space, $W\subset V$, $v+w \in W$ for every $v,w\in W$ and $\alpha u\in W$ for every $u\in W$ and every $\alpha \in \Bbb F$. Finally, suppose that you proved that $x\in W$ for some $x\in V$. We must have $\beta x\in W$ for every $\beta\in \Bbb F$, in particular for $\beta =0$ we get $0=0x\in W$.
This shows that no matter what you can prove to be inside $W$, if $W$ is closed under scalar multiplication and addition, then it has to contain $0$. Nevertheless, note that very very often, showing that $0\in W$ is the simplest way to prove that $W\neq \emptyset$.
For your example: Let $V=\Bbb R^2$ and $W=\{(x,y)\in V\mid x+y=0\}$, then $(1,-1)\in W$ and so $W \neq \emptyset$ (but checking that $(0,0)$ might be even more straightforward).
A: Theorem A subset $W$ of a vector space $V$ is a subspace if and only if (1) $W$ is not empty and, (2) for all scalars $\alpha$, $\beta$ and all $x,y\in W$, $\alpha x+\beta y\in W$.
Remark Note that the property (2) is satisfied by the empty set, which nonetheless isn't a subspace, because of the first requirement.
Proof Only one implication needs to be proved, as the other one is obvious.
So, assume $W$ is not empty and also satisfies property (2). Since $W$ is not empty, there exists $w_0\in W$. Because of property (2), with $\alpha=\beta=0$ and $x=y=w_0$,
$$
0=0w_0+0w_0\in W.
$$
If $x,y\in W$, by property (2) with $\alpha=\beta=1$, we have $x+y\in W$. If $x\in W$ and $\alpha$ is a scalar, use $\beta=0$ and $y=w_0$ in property (2) to conclude that
$$
\alpha x=\alpha x+0w_0\in W.
$$
Therefore $W$ is a subspace. QED
In some cases it's easy to prove that a subset is not empty; so, in order to prove it's a subspace, it's sufficient to prove it's closed under linear combinations.
For instance, let $v_1,\dots,v_n\in V$ and define $\operatorname{Span}\{v_1,\dots,v_n\}$ as the set of all linear combinations of the given vectors. Such a subset is clearly not empty and, since
$$
\alpha(\alpha_1v_1+\dots+\alpha_nv_n)+
\beta(\beta_1v_1+\dots+\beta_nv_n)=
(\alpha\alpha_1+\beta\beta_1)v_1+
\dots+
(\alpha\alpha_n+\beta\beta_n)v_n
$$
is a linear combination, we have that $\operatorname{Span}\{v_1,\dots,v_n\}$ is a subspace of $V$.
