How is the net resistance lesser than the least resistance when resistances are connected in paralle

The current flows through both resistors.
What works for me is to use the analogy of fluid through pipes under pressure.
Imagine two huge tanks, connected by two small pipes of different sizes. The pressures in each of the two tanks are analogous to the two voltages. The pipes are analogous to the resistors. The fluid is analogous to the electrons. The flow (mass flux) of the liquid is analogous to the current.
Each of the pipes has its own linear function of flow as a response to pressure, $F_1=k_{1}p$ and $F_2=k_{2}p$ , where the k's are the flow efficiencies of each pipe for letting liquid flow in response to pressure. A larger pipe will be more efficient at letting liquid through than a smaller pipe. The total flow would be the sum of the flows through each pipe.
$\begin{array}{}\text{(1)}& F=F_1+F_2=k_{1}p+k_{2}p=(k_1+k_2)p=kp\end{array}$
and the total flow efficiency of the two pipe system would be the sum of the two pipe efficiencies,
$\begin{array}{}\text{(2)}& k=k_1+k_2\end{array}$
Flow efficiency for a pipe is analogous to current conductance (ability to conduct current) for a resistor. It would be great if the convention was to use conductance, but we typically describe components with the attribute "resistance", which is the inverse of conductance.
In this scheme, inverses of the pipe efficiencies, $\frac{1}{k_1}$ and $\frac{1}{k_2}$, are analogous to the resistances, $R_1$ and $R_2$, of each of the resistors. Conversely, inverse of the resistances, $\frac{1}{R_1}$ and $\frac{1}{R_2}$, of each of the resistors are analogous to the pipe efficiencies, $k_1$ and $k_2$. Therefore, by analogy to Eq. (2),
$\begin{array}{}\text{(3)}& \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}\end{array}$
and by analogy to Eq. (1),
$\begin{array}{}\text{(4)}& I=I_1+I_2=(\frac{1}{R_1})V+(\frac{1}{R_2})V=(\frac{1}{R_1}+\frac{1}{R_2})V=\frac{V}{R}\end{array}$
I know you were not interested in rederiving the formula, put others may want that.
To summarize in words, not formulae: in the plumbing analogy, it is easy to see that the flow between tanks would not just be the flow through the big pipe, but both. And in the circuit, the current would not just be the current through the "easier" resistor, but both. The easier resistor (big pipe) being the one with smaller resistance.