Let $X_n$ be a symmetric random walk on $\mathbb{Z}$ with $X_0 = 0$. Let $b > 0 > a$ and define the stopping times $$ \tau_a = \min \{ n : X_n = a\}, \ \tau_b = \min \{ n : X_n = b \}$$ and the desired stopping time by $$ \tau = \min \{ \tau_a, \tau_b \} \ .$$ I wish to show that $\mathbb{E}(\tau) < \infty$. Here is my (hopefully correct) proof.
First we note that $X_n$ is a martingale and since $\tau$ is a stopping time $Y_n = X_{\min \{ \tau, n \}}$ is also a martingale. Now note that $$ \mathbb{E}(|Y_n|^2) \leq \max \{ |a|^2, |b|^2 \} \ .$$ This implies that $Y_n$ is uniformly $L^2$ bounded and thus by Doob's convergence theorem there exists a random variable $Y_\infty \in L^2$ such that $$ \lim_{n \to \infty} Y_n = Y_\infty, \ \text{a.s.}$$ and we have convergence in $L^2$.
Next recall that $Z_n = Y_n^2 - \min \{\tau, n\}$ is also a martingale. We have $$ \mathbb{E}(Z_n) = \mathbb{E}(Z_0) = \mathbb{E}(Z_0^2) - \mathbb{E}(0) = 0 \ .$$ So in particular $$ \mathbb{E}(Z_n) = 0 \iff \mathbb{E}(Y_n^2) = \mathbb{E}(\min \{\tau, n\}) \ .$$ Now note that $Y_n^2$ satisfies the conditions of the reverse Fatou lemma as it is uniformly bounded for all $n$ by an integrable function. So in particular we have $$ \limsup_{n \to \infty} \mathbb{E}(\min \{ \tau, n \}) = \limsup_{n \to \infty} \mathbb{E}(Y_n^2) \leq \mathbb{E}( \lim_{n \to \infty} Y_n^2) = \mathbb{E}(Y_\infty^2) < \infty \ .$$ Now assume that $\mathbb{P}(\tau = \infty) > 0$. Then we have $$ \mathbb{E}(\min \{ \tau, n \}) = \mathbb{E}(\min \{ \tau, n \} \chi (\tau < \infty)) + \mathbb{E}(\min \{ \tau, n \} \chi (\tau = \infty)) = \mathbb{E}(\min \{ \tau, n \} \chi (\tau < \infty)) + n \mathbb{P}(\tau = \infty) \ .$$ So we have $$ \limsup_{n \to \infty} n \mathbb{P}(\tau = \infty) \leq \limsup_{n \to \infty} \mathbb{E}(\min \{ \tau, n \}) < \infty \ .$$ Which is a contradiction so we conclude that $\mathbb{P}(\tau = \infty) = 0$. Now finally since $\min \{ \tau, n\}$ satisfies the regular Fatou's lemmas conditions we have $$ \mathbb{E}(\tau) = \mathbb{E}(\liminf_{n \to \infty} \min \{ \tau, n \}) \leq \liminf_{n \to \infty} \mathbb{E}(\min \{ \tau, n \}) \leq \limsup_{n \to \infty} \mathbb{E}(\min \{ \tau, n \}) < \infty \ .$$ So we have $\mathbb{E}(\tau) < \infty$ along with $\mathbb{P}(\tau < \infty) = 1$.
I noticed that most proofs of this fact rely on more "hands on" method which involve bounding the expectation $\mathbb{E}(\tau)$ directly with some arguments about successful sequences and such.
Is my proof correct?