All three results are true.

Beginning with the third, it suffices to show that the set of $n\times n$-matrices in $\mathbb{C}_p$ whose characteristic polynomials have distinct roots is dense since these will certainly be diagonalizable.

Now the characteristic polynomial of a matrix may be described as a polynomial of degree $n$ whose coefficients are themselves polynomial in the entries of the matrix (using the usual definition of the characteristic polynomial of $A$ as the determinant of $tI-A$). It follows easily that the (formal) derivative of the characteristic polynomial also has coefficients that are polynomial in the entries of the matrix. Finally using the theory of resultants http://en.wikipedia.org/wiki/Resultant one may find a single polynomial $f$ in the matrix coefficients so that $f(a_{11},a_{12},\ldots,a_{nn})\neq 0$ for $A=(a_{ij})$ if and only if the characteristic polynomial of $A$ and its derivative have no roots in common. This last happens precisely if the characteristic polynomial has distinct roots.

To summarise the previous two paragraphs we have seen that there exists a single polynomial $f$ in $n^2$ variables such that a matrix $(a_{ij})$ is diagonalizable if and only if $f(a_{ij})\neq 0$. In fact, this works over any algebraically closed field since $f$ actually has coefficients in the integers.

It is now easy to see that $f^{-1}(\mathbb{C}_p\backslash 0)$ is an open dense subset of the $n\times n$ matrices with coefficients in $\mathbb{C}_p$ (it is even Zariski dense) as required.

The second follows easily from the third since any neighbourhood of a diagonal matrix will contain an invertible (and diagonal) matrix or from the comments.

The first has already been dealt with in the comments.