[proofplan]
Apply first-order compactness to the $L^*$-theory $\operatorname{Diag}_{\mathrm{el}}(M) \cup \Gamma$. A model of this theory interprets all constants $c_a$ and all additional symbols of $L^*$. The elementary diagram forces the interpretation map $a \mapsto c_a^{N^*}$ to preserve and reflect every $L$-formula with parameters from $M$, hence gives an elementary embedding of $M$ into the $L$-reduct of $N^*$. Finally, the same model satisfies every sentence in $\Gamma$.
[/proofplan]
[step:Use compactness to obtain a model of the full expanded theory]
Define the $L^*$-theory
\begin{align*}
T := \operatorname{Diag}_{\mathrm{el}}(M) \cup \Gamma.
\end{align*}
By hypothesis, every finite subset of $T$ has an $L^*$-model. Therefore, by the [Compactness Theorem](/theorems/2748) (citing a result not yet in the wiki: Compactness Theorem), there exists an $L^*$-structure $N^*$ such that
\begin{align*}
N^* \models T.
\end{align*}
In particular,
\begin{align*}
N^* \models \operatorname{Diag}_{\mathrm{el}}(M)
\qquad\text{and}\qquad
N^* \models \Gamma.
\end{align*}
[guided]
We package all requirements into one first-order theory:
\begin{align*}
T := \operatorname{Diag}_{\mathrm{el}}(M) \cup \Gamma.
\end{align*}
This is a set of $L^*$-sentences because $L^*$ expands $L(M)$, so every sentence of the elementary diagram and every sentence in $\Gamma$ belongs to the same language $L^*$.
The hypothesis says exactly that every finite subset of this theory has a model. The Compactness Theorem states that if every finite subset of a first-order theory is satisfiable, then the whole theory is satisfiable. Applying compactness to $T$, we obtain an $L^*$-structure $N^*$ satisfying
\begin{align*}
N^* \models T.
\end{align*}
Since $T$ is the union of $\operatorname{Diag}_{\mathrm{el}}(M)$ and $\Gamma$, this means
\begin{align*}
N^* \models \operatorname{Diag}_{\mathrm{el}}(M)
\qquad\text{and}\qquad
N^* \models \Gamma.
\end{align*}
The first part will build the elementary copy of $M$; the second part records that all additional requirements imposed by $\Gamma$ are satisfied in the same expanded structure.
[/guided]
[/step]
[step:Define the candidate elementary embedding using the named constants]
Let $N$ be the $L$-reduct of $N^*$. Define the map
\begin{align*}
j: M &\to N \\
a &\mapsto c_a^{N^*}.
\end{align*}
This map is injective. Indeed, if $a,b \in M$ and $a \neq b$, then the $L(M)$-sentence $c_a \neq c_b$ belongs to $\operatorname{Diag}_{\mathrm{el}}(M)$, because it is true in the canonical $L(M)$-expansion of $M$. Since $N^* \models \operatorname{Diag}_{\mathrm{el}}(M)$, we have
\begin{align*}
c_a^{N^*} \neq c_b^{N^*}.
\end{align*}
Thus $j(a) \neq j(b)$ whenever $a \neq b$.
[guided]
The constants in the language $L(M)$ are designed to name the elements of $M$. Since $N^*$ is an $L^*$-structure and $L^*$ contains $L(M)$, each symbol $c_a$ has an interpretation $c_a^{N^*}$ in the universe of $N^*$.
Let $N$ be the $L$-reduct of $N^*$. This means that $N$ has the same underlying set as $N^*$ but remembers only the symbols from $L$. We define
\begin{align*}
j: M &\to N \\
a &\mapsto c_a^{N^*}.
\end{align*}
We must first check that this map does not collapse two different elements of $M$. Suppose $a,b \in M$ with $a \neq b$. In the canonical $L(M)$-expansion of $M$, the constants $c_a$ and $c_b$ are interpreted as distinct elements, so the sentence $c_a \neq c_b$ is true there. Hence
\begin{align*}
c_a \neq c_b \in \operatorname{Diag}_{\mathrm{el}}(M).
\end{align*}
Because $N^*$ satisfies the elementary diagram, it satisfies this sentence:
\begin{align*}
N^* \models c_a \neq c_b.
\end{align*}
Equivalently,
\begin{align*}
c_a^{N^*} \neq c_b^{N^*}.
\end{align*}
Therefore $j(a) \neq j(b)$, so $j$ is injective.
[/guided]
[/step]
[step:Show that the constant map preserves every formula with parameters from $M$]
Let $\varphi(x_1,\dots,x_n)$ be an $L$-formula, and let $a_1,\dots,a_n \in M$. Consider the $L(M)$-sentence
\begin{align*}
\varphi(c_{a_1},\dots,c_{a_n}).
\end{align*}
If
\begin{align*}
M \models \varphi(a_1,\dots,a_n),
\end{align*}
then $\varphi(c_{a_1},\dots,c_{a_n}) \in \operatorname{Diag}_{\mathrm{el}}(M)$, so
\begin{align*}
N^* \models \varphi(c_{a_1},\dots,c_{a_n}).
\end{align*}
Since $\varphi$ is an $L$-formula and $N$ is the $L$-reduct of $N^*$, this is equivalent to
\begin{align*}
N \models \varphi(j(a_1),\dots,j(a_n)).
\end{align*}
Conversely, if
\begin{align*}
M \not\models \varphi(a_1,\dots,a_n),
\end{align*}
then
\begin{align*}
M \models \neg \varphi(a_1,\dots,a_n),
\end{align*}
so $\neg\varphi(c_{a_1},\dots,c_{a_n}) \in \operatorname{Diag}_{\mathrm{el}}(M)$. Hence
\begin{align*}
N^* \models \neg\varphi(c_{a_1},\dots,c_{a_n}),
\end{align*}
and therefore
\begin{align*}
N \not\models \varphi(j(a_1),\dots,j(a_n)).
\end{align*}
Thus, for every $L$-formula $\varphi(x_1,\dots,x_n)$ and every tuple $(a_1,\dots,a_n) \in M^n$,
\begin{align*}
M \models \varphi(a_1,\dots,a_n)
\iff
N \models \varphi(j(a_1),\dots,j(a_n)).
\end{align*}
[guided]
To prove that $j$ is elementary, we must show that it preserves and reflects truth of all $L$-formulas with parameters from $M$. Let $\varphi(x_1,\dots,x_n)$ be an arbitrary $L$-formula, and choose elements $a_1,\dots,a_n \in M$.
The corresponding $L(M)$-sentence is
\begin{align*}
\varphi(c_{a_1},\dots,c_{a_n}).
\end{align*}
If this formula is true of the tuple $(a_1,\dots,a_n)$ in $M$, then this sentence is true in the canonical expansion of $M$ by the constants $c_a$. By the definition of the elementary diagram, it belongs to $\operatorname{Diag}_{\mathrm{el}}(M)$. Since $N^*$ satisfies the elementary diagram, we get
\begin{align*}
N^* \models \varphi(c_{a_1},\dots,c_{a_n}).
\end{align*}
Because $\varphi$ uses only symbols from $L$, its truth is unchanged when passing from the expanded structure $N^*$ to its $L$-reduct $N$. Therefore
\begin{align*}
N \models \varphi(j(a_1),\dots,j(a_n)).
\end{align*}
For the reverse implication, suppose $\varphi(a_1,\dots,a_n)$ is not true in $M$. Then its negation is true:
\begin{align*}
M \models \neg \varphi(a_1,\dots,a_n).
\end{align*}
Thus the sentence
\begin{align*}
\neg\varphi(c_{a_1},\dots,c_{a_n})
\end{align*}
belongs to $\operatorname{Diag}_{\mathrm{el}}(M)$. Since $N^*$ satisfies the elementary diagram,
\begin{align*}
N^* \models \neg\varphi(c_{a_1},\dots,c_{a_n}),
\end{align*}
and again reducing to the $L$-structure $N$ gives
\begin{align*}
N \not\models \varphi(j(a_1),\dots,j(a_n)).
\end{align*}
Combining the two directions, we have shown that for every $L$-formula $\varphi(x_1,\dots,x_n)$ and every tuple $(a_1,\dots,a_n) \in M^n$,
\begin{align*}
M \models \varphi(a_1,\dots,a_n)
\iff
N \models \varphi(j(a_1),\dots,j(a_n)).
\end{align*}
This is exactly the Tarski criterion for $j$ to be an elementary embedding.
[/guided]
[/step]
[step:Identify $M$ with its elementary image and retain the additional requirements]
The preceding step shows that $j: M \to N$ is an elementary embedding. Since $j$ is injective, its image $j(M) \subset N$ is an elementary copy of $M$ inside the $L$-structure $N$. Identifying each $a \in M$ with $j(a) = c_a^{N^*}$, we may regard $M$ as an elementary substructure of $N$ and write
\begin{align*}
N \succeq M.
\end{align*}
Moreover, from the construction of $N^*$ we already have
\begin{align*}
N^* \models \Gamma.
\end{align*}
Thus the expanded structure $N^*$ realizes all additional constants, relations, functions, and sentence requirements imposed by $\Gamma$, while its $L$-reduct contains an elementary copy of $M$.
[/step]
