Consider the $n\times n$ matrices given by $A_n=1/n*I_{n\times n}$. For every finite $n$ each $A_n$ has rank $n$ but the limit is the zero matrix, and has rank zero. So the limit of a sequence of rank $n$ matrices need not be of rank $n$. Now I am considering the reverse question.
Consider a sequence $A_k \in \mathbb{R}^{m \times n}$ (or $\mathbb{C}^{m \times n}$). Is it possible to construct a sequence such that each $A_k$ has some rank $d \le \min(m,n)$ but: $\operatorname{rank} \left( \lim \limits _{k \to \infty} A_n \right) >d$?
If not how would one go about proving that matrices cannot jump rank?
You can generalize the proof that the set of singular matrices is closed, by considering the set of matrices of rank less than $d$ as the set $f^{-1}({0})$, for $f\colon \mathbb{R}^{m\times n} \to \mathbb{R}$ defined by $$ f(A) = \sum_{A'} \lvert\det(A')\rvert $$ where the sum if over all $(d+1)\times (d+1)$ sub-matrices. Then $f$ is continuous, and $f(A)$ is non-zero if and only if $A$ has rank at least $d+1$.