Observe that in the original circuit the resistors R6 and R7 are short as they have a wire connecting their ends. You can just replace them with a short. Replacing it with a short you are left with R4 and R5 in parallel. It should be fairly easy to solve from here. It is
$$ R3 || (R2 + (R4 || R5)) = 0.6 \Omega $$.
Let's try a very simple approach.
The collection of resistors will have a maximum value resistor \$R_{max}\$, and a minimum value resistor \$R_{min}\$. By definition $$R_{min} <= R_{max}$$ the equality is true if all resistors are equal, otherwise the inequality is true
The series connection will contain \$R_{max}\$, plus at least one other series resistor, which will increase its resistance, so $$R_s > R_{max}$$
The parallel connection will contain \$R_{min}\$, plus at least one other parallel resistor which will decrease its resistance, so $$R_p < R_{min}$$
as $$R_p < R_{min} <= R_{max} < R_s$$ therefore $$R_p < R_s$$
Notes
(1) We assume all the resistances are non-negative. This is true if they are all passive. There are active networks that can exhibit a negative resistance, but they are not relevant to this question.
(2) Resistance is voltage/current. The total resistance of any two resistors in series will have a resistance greater than either component. The total resistor will have a voltage equal to the sum of that over each component, with all currents being equal.
(3) The total resistance of any two resistors in parallel will have a resistance lower than either component. The total resistor will have a current equal to the sum of that through each component, with all voltages being equal.
Best Answer
Your equation is correct - the verbal description is a bit confusing.
For the Thevenin equivalent source resistance, replace the voltage source(s) with shorts and current source(s) with opens and analyze the resistance looking into the output.