I am totally stuck at this problem.... I can show that $P_1(U)$ must be $V_1$ or $0$ and $P_2(U)$ must be $V_2$ or $0$.
The problem is that what contradiction can I get when $P_1(U)=V_1$ and $P_2(U)=V_2$? I tried everything including Schur's lemma but have only failed and it is unbearably frustrating.....
Could anyone please help me?
If you don't assume that $U$ is irreducible, and merely assume that $U$ is neither $\{0\}$ nor all of $V_1 \oplus V_2$, then:
Let $U' \subseteq U$ be an irreducible subrepresentation. Then if $P_1(U') = V_1$, then by Schur's lemma, you have $U' \simeq V_1$. Similarly $P_2(U') = V_2$ implies $U' \simeq V_2$. Since $V_1$ and $V_2$ are nonisomorphic by assumption, you can't have both happen simultaneously.
Then by linear algebra with $U' \neq \{0\}$, it follows that $U' = V_1$ or $U' = V_2$. Say $U' = V_i$, and let $j \neq i$ with $\{i, j\} = \{1, 2\}$.
Then use the fact that $U/U' \subseteq (V_1 \oplus V_2)/V_i \simeq V_j$ to conclude that $U/U' =\{0\}$ and therefore $U = U'$.