As @NateEldredge has aleady pointed out, this particular line doesn't make sense and, moreover, the convergence
$$\mathbb{P}(|W_t| \leq 2) \xrightarrow[]{t \to \infty} 0$$
can be proved much easier using the scaling property.
However, here is a way to fix the proof: Obviously,
$$\mathbb{P}(|W_t| \leq 2) = \mathbb{P}(W_t \leq 2)-\mathbb{P}(W_t \leq -2) = \int_{-2}^2 \frac{d}{dx} \mathbb{P}(W_t \leq x) \, dx.$$
Using $W_t \sim N(0,t)$, we get
$$\mathbb{P}(|W_t| \leq 2) \leq \frac{4}{\sqrt{2\pi t}} \sup_{x \in [-2,2]} \exp(-x^2/2t) \leq \frac{4}{\sqrt{2\pi t}}.$$
This gives
$$\mathbb{P}(\tau<t) \geq 1- \mathbb{P}(|W_t| \leq 2) \geq 1- \frac{4}{\sqrt{2\pi t}}.$$
The question has been edited with a more correct solution than the one below. However, I will keep this as is since it captures enough of the intuition.
This is reflection principle. Intuitively, for all paths that cross $b$, and then go below $a$, you can flip the path $b$ onward, and get a path that goes above $2b - a$
As you define, let $T = \inf \{t \geq 0 W_t > b\} $, and note that $W_T$ is a well-defined construct. Note that $W_s$ and $W'_s = W_{t - s} - W_s$ are independent. So, we have that
$$P(M_t > b, W_t < a) = \int_0^t P(T = s, W'_{s} < a - b) ds $$
and we note that $P(T = s, W'_{s} < a - b) = P(T = s)P(W'_{s} < a - b) = P(T = s) P(W'_{s} > b - a) = P(T = s, W'_{s} > b - a)$
by the independence of the stopping time from the future, and the symmetry of gaussians. Note that
$\int_0^t P(T = s, W'_{s} > b - a) = \int_0^t P(T = s, W_t > 2b - a) = P(W_t > 2b - a)$ since $\{ T \in [0,t] \} \subset \{W_t > 2b - a\} $
The result now follows from the symmetry of $W_t$.
Best Answer
Suppose that $\mathbb{P}(\tau=0)=0$, then $$\mathbb{P}(\exists t_0>0\, \forall t \leq t_0\::\: W_t \leq 0)=1.$$ Since $(-W_t)_{t \geq 0}$ is also a Brownian motion, this implies $$\mathbb{P}(\exists t_0>0\, \forall t \leq t_0\::\: W_t \geq 0)=1.$$ Hence, $$\mathbb{P}(\exists t_0>0\, \forall t \leq t_0, t \in \mathbb{Q}\::\: W_t \geq 0)=1.$$ As $\mathbb{P}(W_t=0)=0$ for each $t \geq 0$, this gives $$\mathbb{P}(\exists t_0>0\, \forall t \leq t_0, t \in \mathbb{Q}\::\: W_t > 0)=1,$$
i.e. $\mathbb{P}(\tau=0)=1$, which clearly contradicts our assumption.
Hence, $\mathbb{P}(\tau=0)>0$, and by Blumenthal's 0-1-law we conclude that $\mathbb{P}(\tau=0)=1$.