Union of two subspaces versus intersection of two subspaces

The intersection $A \cap B$ of two sets $A$ and $B$ contains only elements which are in both subsets. To verify that the intersection $W := U \cap V$ of two subspaces $U,V$ is a subspace, you need to check if for any two elements $a,b \in W$ you also have $a+b \in W$, and similarly that for any element $a \in W$ and every scalar $\lambda \in K$ you have $\lambda a \in W$.

You do that by observing that if $a,b \in W$ then $a,b \in U$ and $a,b \in V$ because the intersection contains only elements that are in both sets. Now, since $a,b \in U$ you get $a+b \in U$ because $U$ is a subspace, and simiarly $a+b \in V$ because $V$ is also a subspace. Which then of course means $a+b \in W$, since $a+b$ lies in both of the intersected sets. The same line of reasoning works for the $\lambda a$ case.

Note that this a a very general principle - a lot of mathematical structures have the property that if $A$ and $B$ fulfill the structure's axioms then $A \cap B$ does too. And the proof very often basically works as in the case of subspaces above.

No, the intersection is the set of all vectors that are in both subspaces. In your example, that would only be the vector $(0,0)$ For a less trivial example, in $\mathbb R^3$ let $U$ be all vectors of the form $(x,y,0)$ and $V$ be all vectors of the form $(0,y,z)$. Then $U \cap V$ is all vectors of the form $(0,y,0)$.

No. The intersection of two sets is the set of elements that are in both of them. In your example the intersection is $\{(0,0)\}$. The union is the set of elements that are in one, or the other, or both.