Möbius inversion formula Let $X$ be non-empty set and $S$ be the set of all subsets of $X$, which will be a poset, under subset relation.
Let $\phi \colon S\rightarrow \mathbb{Z}$ be any function and for each $H\in S$, define $\sigma(H)=\sum_{K\leq H}\phi(K)$.
$\mu$ be the Möbius function defined by $\sum_{K\leq H} \mu(H)=\delta_{K,X}$.
How to show that $\sum_{H\leq X} \sigma(H)\mu(H)=\phi(X)$?
(I considered LHS, and substituted for $\sigma(H)$, interchanged the sum, but I couldn't proceed further...)  

As you said,
$$\sum_{H\le X}\sigma(H)\mu(H)=\sum_{H\le X}\sum_{K\le H}\phi(K)\mu(H)=\sum_{K\le X}\phi(K)\tau_X(K)
$$ with $$\tau_X(K)=\sum_{K\le H\le X}\mu(H),
$$
hence all there is to show is that $\tau_X(X)=1$ and $\tau_X(K)=0$ for every $K\subset X$, $K\ne X$. This is how you defined the Möbius function $\mu$, hence the proof is complete.
