A regular element exists by prime avoidance

An element of $R$ is a zero divisor on $M$ exactly when it belongs to an associated prime: one direction follows from an associated element, and the other from the zero-divisor lemma. Thus an $M$-regular element of $I$ exists exactly when $I$ is not contained in the union of the associated primes.

The associated-prime set is finite for a finite module over a Noetherian ring. Finite prime avoidance therefore says that $I$ is contained in that union exactly when it is contained in one member. Combining this with step 1.1 produces a nonzerodivisor $x\in I$ exactly under the displayed condition; $M\mathrm{/}xM\mathrm{\ne }0$ because $xM=M$ would imply $IM=M$. Thus $x$ is regular in the adopted sense, proving both directions.