From Theorem 9.7 in Lie groups, Lie algebras, and representations written by Hall,
the linear map $ i : \mathfrak{g} \rightarrow U(\mathfrak{g})$ is defined by
For all $X$, $Y \in \mathfrak{g}$, $i([X,Y])=i(X)i(Y)-i(Y)i(X)$.
$U(\mathfrak{g})$ is generated by $\{i(X) : X \in \mathfrak{g}\}$.
If $\mathcal{A}$ is an associative algebra with identity and $j: \mathfrak{g} \rightarrow \mathcal{A}$ is a linear map such that $$ j([X,Y])=j(X)j(Y)-j(Y)j(X)$$ for all $X$, $Y \in \mathfrak{g}$. Then there exists a unique algebra homomorphism $\phi : U(\mathfrak{g}) \rightarrow \mathcal{A}$ such that $\phi(1)=1$ and such that $\phi(i(X))=j(X)$ for all $X\in\mathfrak{g}$.
where $\mathfrak{g}$ is any Lie algebra and $U(\mathfrak{g})$ is an associative algebra with identity.
The question is here. How can I show the linear map $i$ is injective when $\mathfrak{g} \subset Mat_n(\mathbb{C})$ without Poincare-Birkhoff-Witt theorem ?
In order to show, I first assume that $i(X)=i(Y)$ and so I induced $$ i([X,Y])=i(X)i(Y)-i(Y)i(X)=0$$ implies that $[X,Y]=XY-YX=0$ from linearity by condition 1. However, I am stuck. How can I prove that $X=Y$ ?
The one information you have about your $\mathfrak{g}$ is that it is contained in some $M_n(\Bbb C)$, so we should use that somehow. Well, $M_n(\Bbb C)$ is an associative algebra with unit, and the inclusion can be written fancier as an injective (!) map
$$j: \mathfrak{g} \hookrightarrow \mathcal A := M_n(\Bbb C).$$
Now use property 3 to conclude that $i$ must be injective.