Electronic – Simulation shows thestery leakage

If they are open-drain lines during initialization and they are pulling high then try a 1k pull-down resistor so that the 1k resistor wins the battle (after all it's probably only fighting against a weak pull-up during initialization). You need to check this.

After the unit has initialized it can drive the line (and transistors) and operate correctly.

Are you putting a resistor in series with the base on each transistor. Something like 1k should be fine.

With the switches closed you would be trying to turn the relay on. As your circuit stands it doesn't work because the emitter voltage could never rise enough to power the relay. Even if it could it means you would be powering the relay most of the time - a waste of energy. Also, if the door/window was opened momentarily as soon as it was closed the alarm would stop.

By adding a second transistor Q2 (PNP) and a push to break switch you can improve your circuit (and get it to work).

The closed switches turn Q1 OFF (base connected to ground). As there is no current flowing through the 1K resistor Q2 is also OFF. No current can flow through the relay. The only current to flow is through R1 so the circuit will only draw 1.2mA.

If any switch is opened Q1 is turned ON and draws a current through R2 and R3. The voltage across R2 will be clamped by the emitter-base of Q2 to about 0.6V and will turn Q2 ON. With Q2 turned ON its collector current will turn the relay ON.

The diode (D1) is connected across the coil of the relay to prevent damaging Q2 from the back emf when the relay is turned OFF.

The relay switch over and closes the circuit between its common and normally open contacts.

The reset switch (push to OPEN) keeps the relay turned ON even if the door/window switches are closed after opening. It LATCHES the relay. Just omit the reset switch if you don't want this feature.