The Moore-Penrose pseudo-inverse is discontinuous

The case $A=0$. $0^+=0$. Let $A_\varepsilon=\operatorname{diag}(\varepsilon,0,\dots,0)$ with $\varepsilon>0$. Then $A_\varepsilon\to0$ but $A_\varepsilon^+=\operatorname{diag}(1/\varepsilon,0,\dots,0)$ has norm $1/\varepsilon\to\infty$. So $A\mapsto A^+$ is not continuous at $0$ (it is not even bounded near $0$).

General rank $r<n$. Write $A=\sum_{i=1}^r\sigma_iu_iv_i^T$ with $\sigma_i>0$, where $(u_i)$ and $(v_i)$ are orthonormal families. Since $r<n$ we can complete the families and pick unit vectors $u_{r+1},v_{r+1}$ orthogonal to the previous ones. Put
$$A_\varepsilon=A+\varepsilon u_{r+1}v_{r+1}^T.$$
This is a singular value decomposition of $A_\varepsilon$ with $\sigma_{r+1}=\varepsilon>0$, so $\operatorname{rank}A_\varepsilon=r+1$ and
$$A_\varepsilon^+=\sum_{i=1}^r\sigma_i^{-1}v_iu_i^T+\varepsilon^{-1}v_{r+1}u_{r+1}^T,\qquad\|A_\varepsilon^+\|=\max(\sigma_r^{-1},\varepsilon^{-1})\ \xrightarrow[\varepsilon\to0]{}\ \infty.$$
But $\|A_\varepsilon-A\|=\varepsilon\to0$ while $A^+$ is a fixed matrix. So $A^+$ is not the limit of $A_\varepsilon^+$.

Remark. The pseudo-inverse is continuous exactly on the set of matrices of *locally constant rank*, i.e. on each stratum $\{\operatorname{rank}=r\}$ and at the invertible matrices (where $A^+=A^{-1}$). This is why numerical rank decisions ("is this singular value zero?") are delicate.