Fix an algebraic closure $K$ of $F$. Note that since $F$ is infinite, $F^m$ is Zariski-dense in $K^m$ for all $m$. In particular, $M_n(F)$ is Zariski-dense in $M_n(K)$ and $GL_n(F)$ is Zariski-dense in $GL_n(K)$ (since $GL_n(K)$ is an open subset of $M_n(K)$ and $GL_n(F)=GL_n(K)\cap M_n(F)$). Also, the subspace topology on $F^m$ from the Zariski topology on $K^m$ is the same as the Zariski topology over $F$ (if $p\in K[x_1,\dots,x_m]$, then the product of the Galois conjugates of $p$ is a polynomial with coefficients in the perfect closure of $F$ that has the same zero set in $F^m$; raising this polynomial to an appropriate power then gives one with coefficients in $F$).
Now let $f:GL_n(K)\times K^n\to M_n(K)$ be the map $f(A,x)=AD(x)A^{-1}$ where $D$ is the diagonal matrix with diagonal entries given by $x\in K^n$. The image of $f$ is the diagonalizable matrices over $K$, which we know is dense in $M_n(K)$. But we also know that $GL_n(F)\times F^n$ is dense in $GL_n(K)\times K^n$, and so since $f$ is continuous, the image of $f$ restricted to $GL_n(F)\times F^n$ is also dense. That is, the matrices which are diagonalizable over $F$ are dense in all of $M_n(K)$, and thus also in $M_n(F)$ since they happen to be contained in $M_n(F)$ (and as remarked above, the Zariski topology on $M_n(F)$ over $F$ is the same as its subspace topology from $M_n(K)$).
