How can lift be less than thrust that is less than weight?
Your misunderstanding lies in your thought that lift is smaller than thrust, while in fact, lift is much larger than thrust.

The lift is provided by the wings. Their purpose is exactly to create a lift force (upwards force) while requiring relatively little thrust (forwards force). How well they do this is expressed by their lift-to-drag ratio (L/D ratio). A modern airplane can have an L/D of 20 or more. This means that for every newton of thrust, you can lift 20 newtons of weight! 

You can think of wings the same way you think of an inclined plane: you need a lot less force to push a car up a gentle hill than up a steep hill. Indeed, if the hill is gentle, most people can push the car up the hill, while most people cannot single-handedly lift a car up. Wings work by pushing air down, and by pushing the air down at a relatively shallow angle (by going forwards real fast), you need less force to go forwards than to lift the plane.



If you want to know more than this 'intuitive' explanation, perhaps I can recommend you this answer of mine. I will present a terse derivation below, with the aim to calculate the L/D (neglection parasitic drag) purely from first principles.

Lift is due to a certain mass flow $\dot{m}$ being given a certain velocity $v$ downwards: $$L=\dot{m}v.$$ This mass flow is due to the wings encountering a certain amount of air; as an approximation, you can think that the airplane only affects a circular 'tube' of air with the diameter equal to its wingspan. The energy per unit of time (power) necessary to impart this downwards momentum equals $$P=\frac{1}{2}\dot{m}v^2.$$

This power is provided by the combination of thrust $T$ and forwards velocity, $$P=Tu.$$ This gives us an expression for the thrust: $$T=\frac{P}{u} = \frac{1}{2}\dot{m}\frac{v^2}{u}.$$ Note that the thrust $T$ equals drag force $D$, so we can now calculate the L/D:

$$\frac{L}{D}=\frac{L}{T} = \frac{\dot{m}v}{\frac{1}{2}\dot{m}\frac{v^2}{u}} = 2\frac{u}{v}. $$

In other words, like I said before, the L/D ratio is exactly how 'shallow' the relative velocity vector of the affected airflow is. It follows that the forwards speed $u$ should be as large as possible to maximize the L/D. However, I only took into account the so-called "lift induced drag" which is purely the horizontal force it takes to create a vertical lift force. In reality, there is also parasitic drag, which scales with $u^2$ and will at some point dominate over the reduction in L/D.