Is the relativistic Doppler effect the only way to show that the universe is expanding? Is there another way to show that the universe is expanding, rather then only relying on the relativistic Doppler effect?
 A: There is at least one very good piece of evidence that the universe is expanding.
Observations of distant quasars are able to probe the state of matter that gathers into clouds along the line of sight to the quasars. These observations are able to estimate what the temperature of the microwave background was, at the epoch at which the light was emitted (i.e. in the distant past).
These observations (and also less precisely using observations of the Sunyaev Zel'dovich effect towards galaxy clusters) suggest that the microwave background had a higher temperature in the past, exactly as predicted by the adaibatic expansion of the universe: i.e. That $ T =T_0 (1+z)$ to precisions of about 1% to beyond $z=3$ and that the universe was smaller and hotter in the past.
References:
Molaro et al. 2002
Avgoustidis et al. 2016
Lopez-Corredoira et al. 2017
A: I am writing this answer to make things clear. By using general relativity, one can prove that a matter-filled universe is at unstable equilibrium point. If you try to use Newtonian theory, that's another picture. And it that picture yes there is also a solution for the static stable universe. 
However, I believe at this stage, there's no point in introducing the Newtonian static ball model as a counter-argument because I am using General Relativity. Not Newtonian mechanics.
By using Friedmann Equations we can write, 
$$H^2 = \frac{8\pi G }{3c^2}\sum_i\varepsilon_i-\frac{\kappa c^2}{R^2a^2}$$
$$\frac{\ddot{a}}{a}=-\frac{4\pi G }{3c^2}\sum_i(\varepsilon_i + 3p_i)$$
For a matter-filled static universe ($\dot{a} = \ddot{a} = 0$) these equations become, 
$$0 = \frac{8\pi G }{3c^2}\varepsilon_m-\frac{\kappa c^2}{R^2a^2}$$
$$0=-\frac{4\pi G }{3c^2}(\varepsilon_m + 3p_m)$$
Which implies $\varepsilon_m + 3p_m=0$ or $\varepsilon_m = -3p_m$. 
We know that this  is impossbile for a collection of gas and stars.
That's the reason why Einstein Introduced the $\Lambda$ or the cosmological constant. If you put an extra term on the acceleration equation such that
$$\frac{\ddot{a}}{a}=-\frac{4\pi G }{3c^2}\varepsilon_m+\frac{\Lambda}{3}$$
you will see that theres no longer need for  $\varepsilon_m = -3p_m$. 
Where $\Lambda = 4\pi G \rho_m$ and the associated constant energy density is $$\varepsilon_{\Lambda} = \frac{c^2}{8\pi G}\Lambda$$.
Let us write the acceleration equation like this, 
$$\frac{\ddot{a}}{a}=-\frac{4\pi G }{3c^2}(\varepsilon_m - 2\varepsilon_{\Lambda})$$
(for $p_{\Lambda} = -\varepsilon_{\Lambda}$)
The energy density of the $\Lambda$ ($\varepsilon_{\Lambda}$) is constant. This implies that if you increase or decrease the matter density, you 'll get either collapsing or an expanding universe. 
