[proofplan] Set $K := \bigcap_{i \in I} N_i$ and verify the defining properties of a submodule of $M$. Since every $N_i$ is a submodule, every $N_i$ contains $0_M$, is closed under addition, is closed under additive inverses, and is closed under scalar multiplication by elements of $R$. Unpacking membership in the intersection transfers each of these properties from every $N_i$ to $K$. [/proofplan]

[step:Show that the intersection is nonempty] Define the subset $K \subset M$ by \begin{align*} K := \bigcap_{i \in I} N_i. \end{align*} For every $i \in I$, the hypothesis $N_i \le M$ implies $0_M \in N_i$. Therefore $0_M \in K$, so $K$ is nonempty. [/step]

[step:Check closure under addition and additive inverses]Let $x,y \in K$. By the definition of $K$, for every $i \in I$ we have $x \in N_i$ and $y \in N_i$. Since $N_i$ is a submodule of $M$, its underlying additive group is a subgroup of the additive group of $M$, so $x+y \in N_i$ and $-x \in N_i$ for every $i \in I$. Hence $x+y \in K$ and $-x \in K$.[/step]

[guided]We now verify the additive subgroup part of the submodule definition. The only point to keep track of is that membership in an intersection means membership in every set in the family. Let $x,y \in K$. Since \begin{align*} K := \bigcap_{i \in I} N_i, \end{align*} the statement $x \in K$ means that $x \in N_i$ for every $i \in I$, and the statement $y \in K$ means that $y \in N_i$ for every $i \in I$. Fix an arbitrary index $i \in I$. Because $N_i \le M$, the set $N_i$ is an additive subgroup of $M$. Therefore it is closed under addition and additive inverses. Applying these two closure properties inside $N_i$ to the elements $x,y \in N_i$, we obtain \begin{align*} x+y \in N_i \end{align*} and \begin{align*} -x \in N_i. \end{align*} The index $i$ was arbitrary, so both $x+y$ and $-x$ lie in every $N_i$. By the definition of intersection, this gives \begin{align*} x+y \in K \end{align*} and \begin{align*} -x \in K. \end{align*} Thus $K$ is closed under addition and additive inverses.[/guided]

[step:Check closure under scalar multiplication] Let $r \in R$ and $x \in K$. For every $i \in I$, the definition of $K$ gives $x \in N_i$. Since $N_i \le M$, it is closed under the left $R$-action on $M$, so $rx \in N_i$ for every $i \in I$. Hence $rx \in K$. [/step]

[step:Conclude that the intersection is a submodule] The subset $K \subset M$ is nonempty, closed under addition, closed under additive inverses, and closed under scalar multiplication by every element of $R$. Therefore $K$ is a submodule of $M$. Since $K = \bigcap_{i \in I} N_i$, this proves \begin{align*} \bigcap_{i \in I} N_i \le M. \end{align*} [/step]