Much of the post is not on the right track. The part that is on the right track is the list of three conditions. You need to show that the collection of all subsets of a (finite) set satisfies these conditions. The verifications will be in each case very short.
For example, is $S$ in our collection? Sure, every subset of $S$ is in our collection, and $S$ is a subset of itself.
Now that the measures are assumed finite I think this is doable. First we wish to show that $\mu_*(A) = \mu(\Omega) - \mu^*(A)$. Let $\epsilon > 0$. Choose $B_i$ so that $\cup_i B_i \subset A$, and $\mu_*(A) \le \mu(\cup_i B_i) + \epsilon$. It follows since $\mu$ is additive that for $C\in \mathcal{A}$, $\mu(C) = \mu(\Omega) - \mu(C^c)$. A picky detail, but additionally there exists $n$ so that $\mu(\cup_i B_i) \le \mu(\cup_{i=1}^n B_i) + \epsilon$. This is needed since $\cup_{i=1}^n B_i \in \mathcal{A}$ ($\mathcal{A}$ is an algebra). With this, and noting that $ A^c \subset \left(\cup_{i=1}^n B_i \right)^c $,
$$
\mu_*(A) \le \mu(\cup_i B_i) + \epsilon \le \mu(\cup_{i=1}^n B_i) + 2\epsilon = \mu(\Omega) - \mu(\left(\cup_{i=1}^n B_i \right)^c) + 2\epsilon \le \mu(\Omega) - \mu^*(A^c) + 2\epsilon.
$$
Starting with a cover $C_i$ such that $A^c \subset \cup_i C_i$ and $\mu(\cup_i C_i) \le \mu^*(A^c) + \epsilon$, one can show very similarly that
$$
\mu^*(A^c) \ge \mu(\Omega) - \mu_*(A) -2\epsilon.
$$
Manipulating a bit and putting the above equations together we have
$$
\mu(\Omega)-\mu^*(A^c) - 2\epsilon \le \mu_*(A) \le \mu(\Omega)-\mu^*(A^c) + 2\epsilon.
$$
Since $\epsilon$ is arbitrary, we have $\mu_*(A) = \mu(\Omega)-\mu^*(A^c)$. Now this implies that the collection of sets $\mathcal{B}$ is closed under the complement. Suppose $A\in \mathcal{B}$ so that $\mu_*(A)= \mu^*(A)$. Then $\mu_*(A^c) =\mu(\Omega)-\mu^*(A)= \mu(\Omega)-\mu_*(A)= \mu^*(A^c)$.
It also came up in the comments to show that $\mathcal{B}$ is closed under intersection. To that end let $\epsilon>0$, and suppose $A_1,A_2 \in \mathcal{B}$. As a result we may choose $B_{i,1},B_{i,2},C_{i,2},C_{i,2}$ with the following properties: 1) $\cup_i B_{i,j} \subset A_j \subset \cup_i C_{i,j}$, $j=1,2$. 2) $\mu( \cup_{i} C_{i,j} / \cup_{i} B_{i,j}) < \epsilon/2$, $j=1,2$. Evidently then $(\cup_i B_{i,1}) \cap (\cup_i B_{i,2}) \subset A_1 \cap A_2 \subset (\cup_i C_{i,1}) \cap (\cup_i C_{i,2})$. Further, it can be shown that $(\cup_{i } C_{i,1} \cap C_{i,2}) /(\cup_{i } B_{i,1} \cap B_{i,2}) \subset (\cup_{i} C_{i,1} / \cup_{i} B_{i,1}) \cup (\cup_{i} C_{i,2} / \cup_{i} B_{i,2}). $ It follows that $\mu ( (\cup_i C_{i,1} \cap \cup_i C_{i,2}) /(\cup_i B_{i,1}) \cap (\cup_i B_{i,2})) < \epsilon$, giving that $\mu^*(A_1 \cap A_2) - \mu_*(A_1\cap A_2) < \epsilon$. Since $\epsilon$ is arbitrary, $\mu^*(A_1 \cap A_2)=\mu_*(A_1\cap A_2)$.
Best Answer
Yes. That is what you need to do, and roughly how you need to do it. But you should polish it a bit more.
As provided, $1\leqslant\alpha\leqslant 1$, and that $P$ and $Q$ are probability measures over the same event space $(\Omega, \mathcal F)$
This means $P$ and $Q$ obey the three axioms, and we must show that $\alpha P+(1-\alpha)Q$ does too.
Axiom One: Probability of an Event.
For any $A\in\mathcal F$, we have $0\leqslant P(A)\leqslant 1$ and $0\leqslant Q(A)\leqslant 1$.
Therefore also $0\leqslant \alpha P(A) + (1-\alpha)Q(A)\leqslant \alpha+(1-\alpha) = 1$
Axiom Two: Probability of the Outcome Set (Sample Space)
We have $P(\Omega)=1$ and $Q(\Omega)=1$.
Therefore also $\alpha P(\Omega)+(1-\alpha) Q(\Omega)=1$.
[ Likewise $P(\emptyset)=0, Q(\emptyset)=0$ implies $\alpha P(\emptyset)+(1-\alpha) Q(\emptyset)=0$. This is not always included among the axioms, but...eh.]
Axiom Three: Additivity of Probability for Unions of Disjoint Events.
For any sequence of disjoint events $(A_i)$, we have $P(\bigcup_i A_i)=\sum_i P(A_i)$ and $Q(\bigcup_i A_i)=\sum_i Q(A_i)$
Therefore also $\alpha P(\bigcup_i A_i)+(1-\alpha)Q(\bigcup_i A_i)=\sum_i\left(\alpha P(A_i)+(1-\alpha)Q(A_i)\right)$
So if $(\Omega,\mathcal F, P)$ and $(\Omega, \mathcal F, Q)$ are probability spaces, then $(\Omega,\mathcal F, [\alpha P +(1-\alpha)Q] )$ is also a probability space.