How to prove Bonferroni inequalities?

A proof is there. The main idea is that this is the integrated version of analogous pointwise inequalities and that, for every $k$, $$ S_k=\mathbb E\left({T\choose k}\right),\qquad T=\sum_{i=1}^n\mathbf 1_{A_i}. $$ Hence the result follows from the stronger inequalities asserting that, for every positive integer $N$, $$ \sum_{i=0}^k(-1)^ia_i,\qquad a_i={N\choose i}, $$ is nonnegative when $k$ is even and nonpositive when $k$ is odd. In turn, this fact follows from the properties that the sequence $(a_i)_{0\leqslant i\leqslant N}$ is unimodal and that $\sum\limits_{i=0}^N(-1)^ia_i=0$.

Here is a self-contained proof that expands on @Did's remarks.

The assertion is that $\Delta_k\le0$ when $k$ is odd, and $\Delta_k\ge0$ when $k$ is even, where $$ \Delta_k:=P\left(\bigcup_{i=1}^n A_i\right) +\sum_{j=1}^k(-1)^j S_j.\tag1 $$ To prove this, first observe that $S_j$ is the expected value of $$ \sum_{i_1 < \cdots <i_j} I(A_{i_1}\cap\cdots\cap A_{i_j}) = {T \choose j}\tag2 $$ where $I(\cdot)$ denotes an indicator random variable and $T$ is the integer-valued random variable $T:=\sum_{i=1}^n I(A_i)$. The reason is that $I(A_{i_1}\cap\cdots\cap A_{i_j})(\omega)=1$ if and only if $\omega$ belongs to each of the $j$ sets $A_{i_1},\ldots,A_{i_j}$. Thus the LHS of (2) counts the number of ways to select $j$ different $A$'s to which $\omega$ belongs, and so does the RHS of (2). (We follow the convention ${a \choose b}=0$ when $a<b$, so (2) holds even when $T<j$.)

From (2) we see that $\Delta_k$ is the expected value of $$ I(\cup A_i) +\sum_{j=1}^k (-1)^j {T\choose j} \stackrel{(3a)}=I(\cup A_i)\left[\sum_{j=0}^k(-1)^j {T\choose j}\right] \stackrel{(3b)}=I(\cup A_i)\left[(-1)^k{T-1\choose k}\right].\tag3 $$ To justify equality (3a), consider the cases $\omega\in\cup A_i$ and $\omega\not\in\cup A_i$. For (3b) we apply (pointwise) an identity about the truncated sum of alternating binomial coefficients. From this last expression we conclude that (3) is a non-positive random variable when $k$ is odd, and a non-negative random variable when $k$ is even, which implies the claimed result.

As a bonus, plug $k=n$ in (3). Since $T\le n$, the bracketed quantity will be zero, which implies $\Delta_n=0$, which is the inclusion-exclusion principle.