Diagonalizable matrices with complex values are dense in set of $n\times n$ complex matrices.

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 $$ d(S^{-1}J'S,A)=d(S^{-1}J'S,S^{-1}JS)\leq m\,d(J',J)<\varepsilon. $$

Let $X$ be the set of diagonalizable matrices in $M_n(\mathbb C)$, and $Y$ the set of those matrices in $M_n(\mathbb C)$ which have $n$ distinct eigenvalues.

As $Y\subset X\subset M_n(\mathbb C)$, it suffices to show that $Y$ is dense in $M_n(\mathbb C)$.

Proof 1. Let $A$ be in $M_n(\mathbb C)$, and let $U$ be a neighborhood of $A$ in $M_n(\mathbb C)$. It suffices to check that $U$ intersects $Y$.

As $A$ is similar to a triangular matrix, we can assume that $A$ is triangular.

The diagonal entries of a triangular matrix being its eigenvalues, there is a diagonal matrix $D$ such that $A+D$ is in $U\cap Y$. QED

Proof 2. If you know the notion of discriminant of a univariate polynomial, you can argue as follows.

The discriminant $d(A)$ of the characteristic polynomial of $A$ being a nonzero polynomial, with complex (in fact integer) coefficients, in the entries of $A$, the set $$ Y=M_n(\mathbb C)\setminus d^{-1}(0) $$ is dense in $M_n(\mathbb C)$. QED