So I have the following definition from the book:
Definition: A matrix Lie group is any subgroup $G$ of $GL(n, \mathbb{C})$ with the following property: If $A_m$ is any sequence of matrices in $G$, and $A_m$ converges to some matrix $A$ then either $A \in G$, or $A$ is not invertible.
I don't understand how this is the same as being a closed subgroup of $GL(n, \mathbb{C})$. The thing that is confusing me is why do we need the statement $A$ is not invertible?
I'm not sure about this, but what about the sequence $A_m = \frac{1}{m}I$ for $G = GL(n, \mathbb{C})$? Note that $A_m \in GL(n, \mathbb{C})$ for every $m$, but in $M(n, \mathbb{C})$, $A_m \to 0$ which is not invertible. Without the caveat that the limiting matrix is allowed to be non-invertible, $GL(n, \mathbb{C})$ would not be considered a matrix Lie group.
I believe you want to know why $G$ being a matrix Lie group implies that it is a closed subgroup of $GL(n, \mathbb{C})$. Suppose $G$ is a matrix Lie group and suppose $A_m$ is a sequence in $G$ which converges to $A \in M(n, \mathbb{C})$, then $A \in GL(n, \mathbb{C})$ or $A \notin GL(n, \mathbb{C})$ (i.e. $A$ is not invertible). If $A \in GL(n, \mathbb{C})$, then $A \in G$ by definition. So $G$ is a subset of $GL(n, \mathbb{C})$ such that for any sequence in $A_m$ converging to $A \in GL(n, \mathbb{C})$, $A \in G$. This is precisely saying that $G$ is a sequentially closed subset of $GL(n, \mathbb{C})$. As $GL(n, \mathbb{C})$ is metrisable, this is equivalent to $G$ being a closed subset of $GL(n, \mathbb{C})$. As $G$ is a subgroup of $GL(n, \mathbb{C})$ by definition, and $G$ is a closed subset of $GL(n, \mathbb{C})$, $G$ is a closed subgroup of $GL(n, \mathbb{C})$.
Consider the case $G=GL(1,\mathbb{C})=\mathbb{C} \setminus \{ 0\}$ and the sequence $\{\frac{1}{n} \}$. Clearly, this sequence is in $G$ (thought of as a subgroup of $G$ for your example), but converges to $0$, which is not invertible. The point is that the sequence converges outside of the general linear group.
