The set of all invertible matrices is a dense open subset of the set of all $n\times n$ matrices. The set of all unitary matrices is a compact subset of all $n\times n$ matrices. These two sets are also groups under multiplication. They are called the general linear group $\GL(n)$ and the unitary group $\U(n)$, respectively.
Solution.
(1). $\GL(n)$ is a dense subset of $\M(n)$, the set of all $n\times n$ matrices. Indeed, by the Schur triangularisation, for each matrix $A$, there exists a unitary $U$ such that $$\bex A=U\sex{\ba{cccc} \vLm_1&&*\\ &\vLm_1&\\ &&\ddots&\\ &&&\vLm_s \ea},\quad \vLm_i=\sex{\ba{ccc} \lm_i&&*\\ &\ddots&\\ &&\lm_i \ea},\quad \lm_1=0,\quad \lm_i \neq 0,\ 2\leq i\leq s. \eex$$ We may just replace the $\lm_1=0$ by $\ve>0$ to get an invertible matrix $B$ such that $\sen{A-B}_2=\ve^2$.
(2). $\GL(n)$ is an open subset of $\M(n)$. In fact, by continuity, $$\bex \det A_n=0,\quad A_n\to A\ra \det A=0. \eex$$
(3). $\U(n)$ is a bounded, closed subset of $\M(n)$.