We first show that for all $t\ge s$,
$$
\pi_t\pi_s=\pi_s\pi_t = \pi_{s}.
$$ We can observe that $H_s\le H_t$, which implies that
$$
\text{Im}\ \pi_s =H_s \le H_t = \ker(1-\pi_t),
$$
and
$$
\text{Im}\ (1-\pi_t) = H_t^\perp \le H_s^\perp = \ker \pi_s.
$$ This implies $(1-\pi_t)\pi_s = \pi_s(1-\pi_t) = 0$, i.e. $\pi_t\pi_s=\pi_s\pi_t = \pi_s$.
It is a direct consequence that for $t\ge s$,
$$
\langle x,\pi_s x\rangle =\|\pi_sx\|^2=\|\pi_s\pi_tx\|^2\le \|\pi_tx\|^2 =\langle x,\pi_tx\rangle,
$$ since $\|\pi_s\|\le 1$ for every $s\ge 0$.
It remains to prove right continuity of $t\mapsto \pi_t x$ for every $x$. Let $x\in H$ and $s\ge 0$ be arbitrarily given. We can find that for $t\ge s$,
$$
f(t) =\|\pi_{t}x-\pi_s x\|^2=\langle x,\pi_{t}x-\pi_s x\rangle \ge 0
$$ is a decreasing function of $t\in[s,\infty)$. Thus there is a finite limit $\lim_{t\downarrow s}f(t)$, which implies for each and every decreasing sequence $t_n \downarrow s$,
$$
\|\pi_{t_n}x-\pi_{t_m}x\|^2 =\langle x,\pi_{t_n}x-\pi_{m} x\rangle =f(t_n)-f(t_m)\xrightarrow{n,m\to\infty} 0,
$$ i.e. $(\pi_{t_n}x)_{n\ge 1}$ forms a Cauchy sequence. If we denote $y=\lim_{n\to\infty}\pi_{t_n}x$, then we can find that
$$
\|\pi_{t_n}x-\pi_s x\|^2=\|\pi_{t_n}x-\pi_s \pi_{t_n}x\|^2\xrightarrow{n\to\infty} \|y-\pi_s y\|^2\ge 0.
$$ Finally, from the fact that $\pi_{t_n}x \in H_t$, $t>s$ for all sufficiently large $n$'s, we can deduce that
$$
y=\lim_{n\to\infty}\pi_{t_n}x \in H_t,\quad \forall t>s,
$$ hence
$$
y\in \bigcap_{t>s}H_t =H_s.
$$ This proves
$$
\|\pi_{t_n}x-\pi_s x\|^2\xrightarrow{n\to\infty} \|y-\pi_s y\|^2=0
$$ for all decreasing sequence $(t_n)_{n\ge 1}$ with $\lim_{n\to\infty} t_n =s$. Since $s$ was arbitrary, the path $t\mapsto \pi_t x$ is right continuous for every $x$.
There's a lot that can be simplified away from this question. Given that you're only considering finitely many of your nested family of closed subspaces, we might as well just assume we have closed subspaces
$$H_1 \subseteq H_2 \subseteq \ldots \subseteq H_k,$$
with projection $\pi_i$ onto $H_i$.
Also, you don't seem to use $\alpha_1$, so I'm going to rename $\alpha_0$ to $\alpha_1$.
Recall that orthogonal projections are self-adjoint and idempotent. Given $i < j$, since $H_i \subseteq H_j$, we have that $\pi_j \pi_i = \pi_i$. Since the projections are self-adjoint, we also get $\pi_i \pi_j = (\pi_j \pi_i)^* = \pi_i$.
Hence,
$$\pi_i \pi _j = \pi_j \pi_i = \pi_i$$
whenever $i < j$. In fact, it holds even when $i = j$.
Using Pythagoras's theorem, it suffices to show that
$$\pi_1 x, (\pi_2 - \pi_1)x, (\pi_3 - \pi_2)x, \ldots, (\pi_k - \pi_{k - 1})x$$
are pairwise orthogonal. If $2 \le i < j$, then
\begin{align*}
\langle (\pi_i - \pi_{i - 1})x, (\pi_j - \pi_{j - 1})x\rangle &= \langle(\pi_j - \pi_{j - 1})(\pi_i - \pi_{i - 1})x, x \rangle\\
&= \langle(\pi_j\pi_i + \pi_{j - 1} \pi_{i - 1} - \pi_j \pi_{i - 1} - \pi_{j - 1} \pi_i)x, x\rangle \\
&= \langle(\pi_i + \pi_{i - 1} - \pi_{i - 1} - \pi_i)x, x\rangle \\
&= 0.
\end{align*}
Also, given $i \ge 2$, we have
$$\langle (\pi_i - \pi_{i - 1}) x, \pi_1 x \rangle = \langle \pi_1(\pi_i - \pi_{i - 1}) x, x \rangle = \langle (\pi_1 - \pi_1) x, x \rangle = 0.$$
The vectors are all orthogonal, and the result you want holds by Pythagoras's theorem.
Best Answer
Actually, (1) is not true.
The projection is characterized by $\pi_0 f \in H_0$, i.e., $\pi_0 f$ is a constant function (with value $c_f$), and $$(\pi_0 f, g) = (f, g) \qquad \forall g \in H_0.$$
Plug in $g = 1$ to obtain $$ \int_\Omega c_f \, \mathrm d\mu = \int_\Omega f \, \mathrm{d}\mu.$$ Thus, $$ c_f = \frac{\int_\Omega f \, \mathrm d\mu}{\mu(\Omega)}.$$