Obviously, we can write
$$M_T = \mathbb{E} \left[ \left( \int_0^T 1 \, dW_t \right) \cdot \left( \int_0^T t \, dW_t \right) \right],$$
i.e. we are interested in the expectation of a product of stochastic integrals. Using the identity
$$a \cdot b = \frac{1}{4} ((a+b)^2-(a-b)^2) \tag{1}$$
for
$$a := \int_0^T 1 \, dW_t \qquad \quad b := \int_0^T t \, dW_t$$
we get
$$M_T = \frac{1}{4} \mathbb{E} \left[ \left(\int_0^T (1+t) \, dW_t \right)^2 \right] - \frac{1}{4} \mathbb{E} \left[ \left(\int_0^T (1-t) \, dW_t \right)^2 \right].$$
Applying Itô's isometry yields
$$\begin{align*} M_T &= \frac{1}{4} \mathbb{E} \left( \int_0^T (1+t)^2 \, dt \right) - \frac{1}{4} \mathbb{E} \left( \int_0^T (1-t)^2 \, dt \right) \\ &= \int_0^T \frac{1}{4} ((1+t)^2-(1-t)^2) \, dt \\ &\stackrel{(1)}{=} \int_0^T t \, dt. \end{align*}$$
In fact, we have shown the following (more general) statement:
Let $f,g \in L^2([0,T] \otimes \mathbb{P})$ be progressively measurable. Then $$\mathbb{E} \left[ \left( \int_0^T f(t) \, dW_t \right) \cdot \left( \int_0^T g(t) \, dW_t \right) \right] = \mathbb{E} \int_0^T g(t) \cdot f(t) \, dt.$$
Note that for $f=g$ this is the (standard version of) Itô's isometry.
I guess that the problem is asking you to use stochastic integral in place of the optional stopping theorem. Notice that the Itô isometry applied to
$$ W_{T\wedge b} - W_{T\wedge a} = \int_{a}^{b} \mathbf{1}_{\{s < T\}} \, \mathrm{d}W_s, \quad (a < b) $$
yields
$$ \Bbb{E}( |W_{T\wedge b} - W_{T\wedge a}|^2 ) = \int_{a}^{b} \Bbb{P}(s < T) \, \mathrm{d}s.$$
From this and $\int_{0}^{\infty} \Bbb{P}(s < T) \, \mathrm{d}s = \Bbb{E}T < \infty$, we find that $\{W_{T\wedge t} : t \geq 0 \}$ is Cauchy in $L^2$ and hence convergent in $L^2$. But since $T$ is finite a.s., we know that $W_{T\wedge t} \to W_T$ a.s. This identifies the $L^2$-limit of $W_{T\wedge t}$ as $W_T$. Putting these altogether, we have
$$\Bbb{E} W_T^2 = \lim_{t\to\infty} \Bbb{E} W_{T\wedge t}^2 = \lim_{t\to\infty} \int_{0}^{t} \Bbb{P}(s < T) \, \mathrm{d}s = \int_{0}^{\infty} \Bbb{P}(s < T) \, \mathrm{d}s = \Bbb{E}T. $$
Finally, since $W_{T \wedge t} = \int_{0}^{t} \mathbf{1}_{ \{ s < T \}} \, \mathrm{d}W_s$ is an $L^2$-martingale which converges to $W_T$ in $L^2$, it also converges in $L^1$ and hence
$$ \Bbb{E}W_T = \lim_{t\to\infty} \Bbb{E}W_{T \wedge t} = 0. $$
Best Answer
Fubini's theorem on $[0,t]\times\Omega$ gives $$\mathbb{E}\left[\int^{t}_{0} W^2_s\, ds\right] =\int_\Omega \int^t_0 W_s^2(\omega)\,ds\,\mathbb{P}(d\omega) =\int^t_0\int_\Omega W_s^2(\omega)\,\mathbb{P}(d\omega)\,ds = \int^{t}_{0} \mathbb{E}[W^2_s]\, ds.$$