$(1)$证明：连接$OE$，
$\because$四边形$ABCD$是正方形，对角线$AC$、$BD$相交于点$O$，$E$为$CD$边的中点，
$\therefore AO=CO,DE=CE,AD$∥$BC$，
$\therefore \angle DAE=\angle CFE$，
在$\triangle DAE$和$\triangle CFE$中，
$\left\{\begin{array}{l}{∠DAE=∠CFE}\\{∠AED=∠FEC}\\{DE=CE}\end{array}\right.$，
$\therefore \triangle DAE$≌$\triangle CFE\left(AAS\right)$，
$\therefore AE=FE$，
$\therefore OE$∥$CF$，$OE=\frac{1}{2}CF$，
$\therefore \triangle EON$∽$\triangle CFN$，
$\therefore \frac{ON}{FN}=\frac{OE}{CF}=\frac{1}{2}$，
$\therefore FN=2ON$.
$(2)$证明：由（1）得$\triangle DAE$≌$\triangle CFE$，
$\therefore FC=AD$，
$\therefore BC=AD$，
$\therefore FB=2AD$，
$\because AD$∥$FB$，
$\therefore \triangle ADM$∽$\triangle FBM$，
$\therefore \frac{AM}{FM}=\frac{AD}{FB}=\frac{1}{2}$，
$\therefore FM=2AM$，
$\therefore \frac{FM}{FA}=\frac{2AM}{2AM+AM}=\frac{2}{3}$，
$\because \frac{FN}{FO}=\frac{2ON}{2ON+ON}=\frac{2}{3}$，
$\therefore \frac{FM}{FA}=\frac{FN}{FO}$，
$\because \angle MFN=\angle AFO$，
$\therefore \triangle MFN$∽$\triangle AFO$，
$\therefore \angle FMN=\angle FAO$，
$\therefore MN$∥$AO$，
$\therefore MN$∥$OC$，
$\therefore \triangle DMN$∽$\triangle DOC$，
$\therefore \frac{DM}{DO}=\frac{DN}{DC}$，
$\because DO=BO=\frac{1}{2}BD$，$OC=OA=\frac{1}{2}AC$，且$BD=AC$，
$\therefore DO=OC$，
$\because DC=BC$，
$\therefore \frac{DM}{OC}=\frac{DN}{BC}$.
$(3)$$\because AB=BC=4$，$\angle ABC=90^{\circ}$，
$\therefore AC=\sqrt{A{B}^{2}+B{C}^{2}}=\sqrt{{4}^{2}+{4}^{2}}=4\sqrt{2}$，
$\therefore AO=\frac{1}{2}AC=2\sqrt{2}$，
$\because \triangle MFN$∽$\triangle AFO$，
$\therefore \frac{MN}{AO}=\frac{FM}{FA}=\frac{2}{3}$，
$\therefore MN=\frac{2}{3}AO=\frac{2}{3}\times 2\sqrt{2}=\frac{4\sqrt{2}}{3}$，
$\therefore MN$的长为$\frac{4\sqrt{2}}{3}$.

$(1)$证明：连接$OE$，
$\because$四边形$ABCD$是正方形，对角线$AC$、$BD$相交于点$O$，$E$为$CD$边的中点，
$\therefore AO=CO,DE=CE,AD$∥$BC$，
$\therefore \angle DAE=\angle CFE$，
在$\triangle DAE$和$\triangle CFE$中，
$\left\{\begin{array}{l}{∠DAE=∠CFE}\\{∠AED=∠FEC}\\{DE=CE}\end{array}\right.$，
$\therefore \triangle DAE$≌$\triangle CFE\left(AAS\right)$，
$\therefore AE=FE$，
$\therefore OE$∥$CF$，$OE=\frac{1}{2}CF$，
$\therefore \triangle EON$∽$\triangle CFN$，
$\therefore \frac{ON}{FN}=\frac{OE}{CF}=\frac{1}{2}$，
$\therefore FN=2ON$.
$(2)$证明：由（1）得$\triangle DAE$≌$\triangle CFE$，
$\therefore FC=AD$，
$\therefore BC=AD$，
$\therefore FB=2AD$，
$\because AD$∥$FB$，
$\therefore \triangle ADM$∽$\triangle FBM$，
$\therefore \frac{AM}{FM}=\frac{AD}{FB}=\frac{1}{2}$，
$\therefore FM=2AM$，
$\therefore \frac{FM}{FA}=\frac{2AM}{2AM+AM}=\frac{2}{3}$，
$\because \frac{FN}{FO}=\frac{2ON}{2ON+ON}=\frac{2}{3}$，
$\therefore \frac{FM}{FA}=\frac{FN}{FO}$，
$\because \angle MFN=\angle AFO$，
$\therefore \triangle MFN$∽$\triangle AFO$，
$\therefore \angle FMN=\angle FAO$，
$\therefore MN$∥$AO$，
$\therefore MN$∥$OC$，
$\therefore \triangle DMN$∽$\triangle DOC$，
$\therefore \frac{DM}{DO}=\frac{DN}{DC}$，
$\because DO=BO=\frac{1}{2}BD$，$OC=OA=\frac{1}{2}AC$，且$BD=AC$，
$\therefore DO=OC$，
$\because DC=BC$，
$\therefore \frac{DM}{OC}=\frac{DN}{BC}$.
$(3)$$\because AB=BC=4$，$\angle ABC=90^{\circ}$，
$\therefore AC=\sqrt{A{B}^{2}+B{C}^{2}}=\sqrt{{4}^{2}+{4}^{2}}=4\sqrt{2}$，
$\therefore AO=\frac{1}{2}AC=2\sqrt{2}$，
$\because \triangle MFN$∽$\triangle AFO$，
$\therefore \frac{MN}{AO}=\frac{FM}{FA}=\frac{2}{3}$，
$\therefore MN=\frac{2}{3}AO=\frac{2}{3}\times 2\sqrt{2}=\frac{4\sqrt{2}}{3}$，
$\therefore MN$的长为$\frac{4\sqrt{2}}{3}$.