In the completed circuit you still did not meet the Zin requirements, because the resistance (Beta x RE1) is in parallel with RB1 and RB2 as well.
50K is a large input impedance for BJT as well as a small battery voltage.
If the supply voltage was higher than the Zin could be met with BJT, but 3V. is too small to get that kind of Zin.
With the supply of 3v. your using ,You could use a FET in the input, or make a BJT buffer stage, to handle the Zin.
However the FET would be pushing it with a smallish battery.
Probably a BJT buffer of some sort to get a required Zin.
Something like this
Zin =~60K using beta of 50 for both transistors
Gain =~5
Zout =~5K
The reason for higher Zin is the 2 transistor emitter resitances multiplied by Beta.
Even assuming a small value of 50 for beta still gives significant input resistances.