I cannot really follow the reasoning you are hinting in your question, but here's my take:
To talk about density you need a topology. Since $M_n(\mathbb{C})$, the space of complex $n\times n$ matrices is finite-dimensional, a very natural notion of convergence is entry-wise; so we can consider the metric
$$
d(A,B)=\max\{ |A_{kj}-B_{kj}|\ : k,j=1,\ldots,n\}, \ \ \ A,B\in M_n(\mathbb{C}).
$$
It is not hard to check that for any matrix $C$,
$$
d(CA,CB)\leq d(A,B)\,\sum_{k,j=1}^n |C_{kj}|,
$$
and the same inequality holds for multiplication on the right (this will be used in the last inequality below).
Now take any $A\in M_n(\mathbb{C})$. Let $J$ be its Jordan canonical form; then there exists a non-singular matrix $S$ such that $J=SAS^{-1}$. Fix $\varepsilon>0$. Let
$$
m=\left(\sum_{k,j=1}^n |S_{kj}|\right)\,\left(\sum_{k,j=1}^n |(S^{-1})_{kj}|\right)
$$
Now, the matrix $J$ is upper triangular, so its eigenvalues (which are those of $A$) are the diagonal entries. Let $J'$ be the matrix obtained from $J$ by perturbing the diagonal entries of $J$ by less than $\varepsilon/m$ in such a way that all the diagonal entries of $J'$ are distinct.
But now $J'$ is diagonalizable, since it has $n$ distinct eigenvalues. And $d(J,J')<\varepsilon/m$. Then $S^{-1}J'S$ is diagonalizable and
$$
d(S^{-1}J'S,A)=d(S^{-1}J'S,S^{-1}JS)\leq m\,d(J',J)<\varepsilon.
$$
If $\|A-I\|<1$ you can always define a square root with the Taylor series of $\sqrt{1+u}$ at $0$:
$$
\sqrt{A}=\sqrt{I+(A-I))}=\sum_{n\geq 0}\binom{1/2}{n}(A-I)^n.
$$
If $A$ is moreover symmetric, this yields a symmetric square root.
More generally, if $A$ is invertible, $0$ is not in the spectrum of $A$, so there is a $\log$ on the spectrum. Since the latter is finite, this is obviously continuous. So the continuous functional calculus allows us to define
$$
\sqrt{A}:=e^\frac{\log A}{2}.
$$
By property of the continuous functional calculus, this is a square root of $A$.
Now note that $\log$ coincides with a polynomial $p$ on the spectrum (by Lagrange interpolation, for instance). Note also that $A^t$ and $A$ have the same spectrum. Therefore
$$
\log(A^t)=p(A^t)=p(A)^t=(\log A)^t.
$$
Taking the Taylor series of $\exp$, it is immediate to see that $\exp(B^t)=\exp(B)^t$.
It follows that if $A$ is symmetric, then our $\sqrt{A}$ is symmetric.
Now if $A$ is not invertible, certainly there is no log of $A$ for otherwise
$$
A=e^B\quad\quad\Rightarrow \quad 0=\mbox{det}A=e^{\mbox{Tr}B}>0.
$$
I am still pondering the case of the square root.
Best Answer
Case 1: Invertible Matrices
For each $A \in M_{n\times n}(\mathbb{C})$, let us consider the characteristic polynomial \begin{align} p_A(z) = \det(zI-A). \end{align} Then it is clear that $A$ is not invertible if and only if zero is a root of of $p_A(z)$. However, if we perturb $A$ by a diagonal matrix, i.e. \begin{align} B= A+\epsilon I \end{align} then we see that $B$ is invertible since the characteristic polynomial of $B$ is $p_B(z) = p_A(z-\epsilon)$ no longer has zero as its root for any $\epsilon \neq 0$. Hence the class of invertible matrices are dense in $M_n(\mathbb{C})$. (In fact, the set of invertible matrices is actually open in $M_n(\mathbb{C})$. )
Case 2: Unitary Matrices
A quick way to see that this class is not dense is to consider the following matrix \begin{align} A = \begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix} \end{align} which is not unitary. In particular, for all unitary matrix $U$ we have \begin{align} \operatorname{Tr}|A-U|\geq \big|\operatorname{Tr}|A|-\operatorname{Tr}|U| \big|= 1. \end{align} Since all norms are equivalent on finite dimensional vector spaces, we have that the class of unitary matrices can't be dense in $M_n(\mathbb{C})$.
Case 3: Symmetric Matrices
A quick way to see that this class is not dense is to consider the matrix \begin{align} C = \begin{pmatrix} 1 & 100\\ 0 & 1 \end{pmatrix} \end{align} which is clearly not symmetric. Now, for any symmetric matrix $S$, we have \begin{align} C-S = \begin{pmatrix} 1-a & 100-b\\ -b & 1-c \end{pmatrix}. \end{align} If we take the Frobenius norm of $C-S$, we get \begin{align} \|C-S\|_F = \sqrt{|1-a|^2+|1-c|^2+b^2+|100-b|^2}\geq \sqrt{b^2+|100-b|^2} \geq \frac{100}{\sqrt{2}}. \end{align} Hence the class of symmetric matrices is not dense in $M_n(\mathbb{C})$.
Case 4: Diagonalizable Matrices Let $A$ be an arbitrary matrix in $M_n(\mathbb{C})$. Let us show that $A$ can be approximated by diagonal matrix.
Using the fact that a matrix is diagonalizable if it has $n$ distinct eigenvalues, we shall construct a $B$ matrix arbitrarily close to $A$ in norm and $B$ has distinct $n$-eigenvalues.
Suppose $p_A(z) =\det(zI-A) =(z-\lambda_1)(z-\lambda_2)\cdots (z-\lambda_n)$ is the characteristic polynomial of $A$ where $|\lambda_1\leq |\lambda_2|\leq \ldots \leq |\lambda_n|$, then consider a diagonal matrix \begin{align} D= \begin{pmatrix} d_1 & 0 & \ldots & 0\\ 0 & d_2 & \ldots & \vdots\\ \vdots & \ldots &\ddots & 0\\ 0 & \ldots & 0 & d_n \end{pmatrix} \end{align} where the $|d_1|<|d_2|<\ldots <|d_n|$. Then by considering the Jordan canocial form of $A$, we have that \begin{align} A+\epsilon D \end{align} has distinct eigenvalues, i.e. diagonaliable for any $\epsilon\neq 0$. Hence the set of diagonalizable matrices are dense in $M_n(\mathbb{C})$. (My explanation of this part is flimsy. Essentially, I'm choosing $d_1, \ldots, d_n$ so that $|\lambda_1+d_1|<|\lambda_2+d_2|<\ldots<|\lambda_n+d_n|$.)