Prove that a conditional probability is a valid probability space

Question

Let $(Ω, P(·))$ denote a probability space and suppose that B is an event in this space where $P(B) > 0$.

Prove that $(Ω, P(·|B))$ is a valid probability space by showing that it satisfies the three axioms:

1. $P(A|B) ≥ 0$ for all events $A ⊆ Ω$.
2. $P(Ω|B) = 1$.
3. For any two mutually exclusive events C and D in $Ω, P(C \cup D|B) = P(C|B) + P(D|B)$.

So far I have:

1. $P(A|B) = \frac{P(A \cap B)}{P(B)}$, since $P(B) > 0$ and $A ⊆ Ω$ (we know that A is non-negative), then $P(A \cap B) ≥ 0.$ Therefore, the numerator $≥ 0$ and denominator $> 0$, so $P(A|B) ≥ 0$
2. $P(Ω|B) = \frac{P(Ω \cap B)}{P(B)}$. I think we can observe that $P(Ω \cap B) = P(B)$ because $B$ is in the space of $Ω$, so $P(Ω|B) = \frac{P(B)}{P(B)} = 1$, but I'm not sure how to show it
3. We know that if two events are mutually exclusive, then $P(C \cup D) = P(C) + P(D)$. From there I'm not sure if we can jump to the fact that $P(C \cup D|B) = P(C|B) + P(D|B)$

I'm not sure if my logic is correct and would greatly appreciate feedback/improvements.