[proofplan]
Let $A$ denote the displayed set of all finite $R$-linear combinations of elements of $S$, with the empty sum included. We first prove that $A$ is a submodule of $M$ by checking that it contains $0_M$, is closed under addition, additive inverses, and scalar multiplication. Since $S \subset A$, the minimality of $\langle S \rangle_R$ gives $\langle S \rangle_R \subset A$. Conversely, every submodule containing $S$ contains all finite $R$-linear combinations of elements of $S$, so $A \subset \langle S \rangle_R$.
[/proofplan]
[step:Define the set of finite linear combinations]
Define $A \subset M$ by
\begin{align*}
A = \left\{\sum_{i=1}^{n} r_i s_i : n \in \mathbb{Z}_{\ge 0},\ r_i \in R,\ s_i \in S \text{ for every } i \in \{1,\dots,n\}\right\}.
\end{align*}
For $n=0$, the expression is by convention the empty sum and equals $0_M$. Hence $0_M \in A$, so $A$ is nonempty.
[/step]
[step:Show that finite linear combinations form a submodule]
We prove that $A \le M$. Let $x,y \in A$. Choose $n,m \in \mathbb{Z}_{\ge 0}$, elements $r_i \in R$ and $s_i \in S$ for $1 \le i \le n$, and elements $t_j \in R$ and $u_j \in S$ for $1 \le j \le m$, such that
\begin{align*}
x = \sum_{i=1}^{n} r_i s_i
\end{align*}
and
\begin{align*}
y = \sum_{j=1}^{m} t_j u_j.
\end{align*}
By concatenating these two finite sums, $x+y$ is again a finite $R$-linear combination of elements of $S$, so $x+y \in A$.
For the additive inverse, the module identity $-(r_i s_i) = (-r_i)s_i$ gives
\begin{align*}
-x = \sum_{i=1}^{n} (-r_i)s_i.
\end{align*}
Thus $-x \in A$. Finally, if $a \in R$, then associativity of the left module action gives
\begin{align*}
ax = \sum_{i=1}^{n} (a r_i)s_i.
\end{align*}
Since $a r_i \in R$ for each $i$, we have $ax \in A$. Therefore $A$ is an additive subgroup of $M$ closed under the left $R$-action, so $A$ is a submodule of $M$.
[guided]
We need to prove that $A$ is itself a submodule, because the generated submodule $\langle S \rangle_R$ is defined as the smallest submodule containing $S$. The set $A$ is nonempty because the case $n=0$ is allowed, and the corresponding empty sum is $0_M$.
Now take arbitrary elements $x,y \in A$. By the definition of $A$, there are integers $n,m \in \mathbb{Z}_{\ge 0}$, coefficients $r_i,t_j \in R$, and elements $s_i,u_j \in S$ such that
\begin{align*}
x = \sum_{i=1}^{n} r_i s_i
\end{align*}
and
\begin{align*}
y = \sum_{j=1}^{m} t_j u_j.
\end{align*}
The sum $x+y$ is obtained by putting the two finite lists of summands together. This produces another finite expression whose coefficients lie in $R$ and whose module elements lie in $S$. Hence $x+y \in A$.
Next, additive inverses stay inside $A$. Since $M$ is a left $R$-module, scalar multiplication is compatible with the additive group structure of $M$, so
\begin{align*}
-(r_i s_i) = (-r_i)s_i
\end{align*}
for each $i$. Therefore
\begin{align*}
-x = \sum_{i=1}^{n} (-r_i)s_i,
\end{align*}
which is again a finite $R$-linear combination of elements of $S$. Hence $-x \in A$.
Finally, let $a \in R$. The left module associativity law says that $a(r_i s_i) = (a r_i)s_i$ for every coefficient $r_i \in R$ and every $s_i \in S$. Therefore
\begin{align*}
ax = \sum_{i=1}^{n} (a r_i)s_i.
\end{align*}
Each product $a r_i$ lies in $R$, so this is another finite $R$-linear combination of elements of $S$. Hence $ax \in A$. We have shown that $A$ is nonempty, closed under addition, closed under additive inverses, and closed under scalar multiplication by elements of $R$. Thus $A$ is a submodule of $M$.
[/guided]
[/step]
[step:Use minimality to prove $\langle S \rangle_R \subset A$]
We first show that $S \subset A$. Let $s \in S$. Since $R$ is unital and $M$ is a unital left $R$-module, $1_R s = s$. Taking $n=1$, $r_1=1_R$, and $s_1=s$, we get $s \in A$. Hence $S \subset A$.
Since $A$ is a submodule of $M$ containing $S$, and $\langle S \rangle_R$ is the smallest submodule of $M$ containing $S$, it follows that
\begin{align*}
\langle S \rangle_R \subset A.
\end{align*}
[/step]
[step:Show every containing submodule contains all finite linear combinations]
Let $L \le M$ be any submodule such that $S \subset L$. We prove that $A \subset L$. Let $x \in A$, and choose $n \in \mathbb{Z}_{\ge 0}$, coefficients $r_i \in R$, and elements $s_i \in S$ for $1 \le i \le n$ such that
\begin{align*}
x = \sum_{i=1}^{n} r_i s_i.
\end{align*}
If $n=0$, then $x=0_M$, and $0_M \in L$ because $L$ is a submodule. If $n \ge 1$, then $s_i \in S \subset L$ for every $i$, so $r_i s_i \in L$ by closure of $L$ under scalar multiplication. Since $L$ is closed under finite addition, the sum $\sum_{i=1}^{n} r_i s_i$ lies in $L$. Thus $x \in L$, and therefore $A \subset L$.
Applying this to $L=\langle S \rangle_R$, which is a submodule containing $S$, gives
\begin{align*}
A \subset \langle S \rangle_R.
\end{align*}
[/step]
[step:Conclude the equality and the empty generating case]
The two inclusions give
\begin{align*}
\langle S \rangle_R = A.
\end{align*}
This is exactly the displayed description of $\langle S \rangle_R$.
If $S=\varnothing$, then no expression with $n \ge 1$ is possible because there are no elements $s_i \in S$. The only allowed finite linear combination is the case $n=0$, namely the empty sum $0_M$. Therefore
\begin{align*}
\langle \varnothing \rangle_R = \{0_M\}.
\end{align*}
[/step]
