Consider a point $P$ such that $APNC$ would be a parallelogram. Then $AP=NC=BM$, $AP\parallel NC\parallel BM$ meaning $APBM$ is a parallelogram. Then $\angle PAK=\angle LBM$. Also $AP=BM$ and $AK=BL$. From this we get that $\triangle PAK=\triangle MBL$. So $PK=ML$. Finally we have $$KN+LM=KN+PK\geq PN=AC\blacksquare$$ Remark: In this types of problems you often want to "combine" the two segments two form a triangle the third side of which equals to your desired result. In this case we formed a triangle $PKN$ with $PK=LM$ and $PN=AC$.

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Solution with vectors: $\overrightarrow{AC}=\overrightarrow{BC}-\overrightarrow{BA}$. Also $$\overrightarrow{LM}=\overrightarrow{BM}-\overrightarrow{BL}=\overrightarrow{NC}-\overrightarrow{BL}$$$$\overrightarrow{KN}=\overrightarrow{BN}-\overrightarrow{BK}=\overrightarrow{BN}-\overrightarrow{LA}$$. Meaning $\overrightarrow{KN}+\overrightarrow{LM}=(\overrightarrow{BN}+\overrightarrow{NC})-(\overrightarrow{LA}+\overrightarrow{BL})=\overrightarrow{BC}-\overrightarrow{BA}$ Now $|\overrightarrow{AC}|=|\overrightarrow{LM}+\overrightarrow{KN}|\leq |\overrightarrow{LM}|+|\overrightarrow{KN}|$