A subspace is just a vector space 'contained' in another vector space. To show that $W \subset V$ is a subspace, we have to show that it satisfies the vector space axioms. However, since $V$ is itself a vector space, most of the axioms are basically satisfied already.

Then, we need only show that $W$ is closed under addition and scalar multiplication. To show that $W$ is closed under addition, we show that for any $w_1,w_2 \in W$, $w_1+w_2 \in W$ as well. To show that $W$ is closed under scalar multiplication, we need to show that for any $\alpha \in \Bbb R$ (assuming you are working with real numbers, you probably are), $\alpha w_1 \in W$.

Example: Show that the solutions $(x,y,z)$ of the equation $ax+ by + cz = 0$ form a subspace of $\Bbb R^3$.

Solution: We will call the set of solutions $S$, and show that $S$ is a subspace of $\Bbb R^3$. It suffices to show that $S$ is closed under addition and scalar multiplication. Suppose $(x_1,y_1,z_1), (x_2,y_2,z_2) \in \Bbb R^3$. Then, $$ ax_1 + by_1 + cz_1 = 0, \quad ax_2+ by_2 + cz_2 = 0. $$ We have that $$ (ax_1+ by_1 + cz_1) + (ax_2+ by_2 + cz_2) = a(x_1+x_2)+ b(y_1+y_2) + c(z_1+z_2) = 0, $$ so $(x_1+x_2,y_1+y_2,z_1+z_2) = (x_1,y_1,z_1) + (x_2,y_2,z_2) \in S$. Thus $S$ is closed under addition. Now, suppose that $\alpha \in \Bbb R$. Then we have that $$ \alpha(ax_1 + by_1 + cz_1) = a(\alpha x_1) + b(\alpha y_1) + c(\alpha z_1) = 0 $$ so $(\alpha x_1, \alpha y_1, \alpha z_1) = \alpha(x_1,y_1,z_1) \in S$. Thus $S$ is closed under scalar multiplication. $S$ is closed under addition and scalar multiplication, therefore $S$ is a subspace of $\Bbb R^3$.