We begin with the following Claim: Claim: If $x,x+1 \in A$, then $x \leq 31$. Furthermore, if $x,x+2 \in A$, then $x \leq 43$. Proof: Easy to see from the given hypothesis $\blacksquare$ Therefore, by partitioning the elements of the set $X=\{1, 2, 3, \ldots, 50\}$ in 10 groups as follows, Group 1: $X_1=\{1,2,\ldots, 31 \}$, Group 2: $X_2=\{32,33 \}$, Group 3: $X_3=\{34,35 \}$, $\ldots$ Group7: $X_7=\{42,43 \}$, Group 8: $X_8=\{44,45,46 \}$, Group 9: $X_9=\{47,48,49 \}$ και Group 10: $X_{10}=\{50\}$, we note that $A$ can contain at most one element from each $X_i$, with $2 \leq i \leq 7$, and at most one element from each $X_i$, with $8 \leq i \leq 9$. In total, we have $|A| \leq 31+(1+1+\ldots+1)+(1+1)+1=31+6+2+1=40$. An example showing that $|A|=40$ is feasible, is the set $\{1,2,\ldots,31,32,34,36,38,40,42,44,47,50 \}$. To check that this set works, note that for all $a,b,c \in A$ with $a<b<c$, we have $\dfrac{1}{a}-\dfrac{1}{c}=(\dfrac{1}{a}-\dfrac{1}{b})+(\dfrac{1}{b}-\dfrac{1}{c})>\dfrac{1}{a}-\dfrac{1}{b},$ so it suffices to check the hypothesis for each two consecutive (when written in increasing order) elements. This is immediate from the Claim.