Expected value of Brownian motion at a time decided by a rate one Poisson process. Situation: 

We have that $\{W_{t},t\geq 0\}$ is a Brownian motion and
  $\{N_{t},t\geq 0\}$ is a Poisson process such that $N_{t}$ follows a
  Poisson distribution with parameter $t$. This process is independent
  from our Brownian motion.

Question: a) Show that $\mathbb{E}[W^{2}_{N_{t}}]=t$
b) For $a\geq 0$ we define the stopping time: $\tau=\inf\{t\geq 0:|W_{t}|>a\}$. Show that this stopping time is finite.
My attempt: a) Keeping in mind that Brownian motion has independent normally distributed increments, I rewrite the problem into:
$\mathbb{E}[W^{2}_{N_{t}}]=\mathbb{E}[(W_{N_{t}}-W_{s}+W_{s})^{2}]$ where $0\leq s<N_{t}$.
Using the Brownian motion properties a few more times, this reduces to $\mathbb{E}[W^{2}_{N_{t}}]=N_{t}$, which looks like the desired answer but obviously is not. I thought of doing the same trick but with $N_{s}$ instead of $s$, but again no luck. This is a trick I learned during lectures but I'm starting to doubt if it works here aswell.
b) Intuitively, this makes obvious sense. Eventually we will cross hit $a$ and thus the stopping is finite. However, I struggle to make this a solid proof. Perhaps I could use the Law of Large Numbers but I'm not sure if it is applicable here.
Any help is appreciated.
 A: Note that $N_t$ is a random variable and therefore the statement "where $0 \leq s < N_t$" doesn't make sense.

$N_t$ takes values in $\mathbb{N}_0$ and because of the independence from $(W_t)_{t \geq 0}$ we have
$$\mathbb{E}(W_{N_t}^2) = \mathbb{E} \left( \sum_{k \geq 0} W_k^2 1_{\{N_t = k\}} \right) = \sum_{k \geq 0} \underbrace{\mathbb{E}(W_k^2)}_{=k} \underbrace{\mathbb{P}(N_t=k)}_{=e^{-t} t^k/k!} = t e^{-t} \sum_{k \geq 1} \frac{t^{k-1}}{(k-1)!} =t.$$
Alternatively, we can use the tower property of the conditional expectation to show that
$$\mathbb{E}(W_{N_t}^2) = \mathbb{E} \big[ \mathbb{E}(W_{N_t}^2 \mid N_t) \big] = \mathbb{E} \big[ \underbrace{\mathbb{E}(W_k^2)}_{=k} \big|_{k=N_t} \big] = \mathbb{E}(N_t)=t.$$

To prove that the stopping time $\tau$ is almost surely finite we can use the optional stopping theorem. Since $M_t := W_t^2-t$ is a martingale, it follows from the optional stopping theorem that $$\mathbb{E}(W_{t \wedge \tau}^2) = \mathbb{E}(t \wedge \tau).$$ Since $|W_{t \wedge \tau}| \leq a$ this gives $$\mathbb{E}(t \wedge \tau) \leq a^2,$$ and now the monotone convergence theorem yields $$\mathbb{E}(\tau) \leq a^2$$ which shows, in particular, that $\tau<\infty$ almost surely.
