In this case, we can give an easy estimate on the tail probability of $T_1$. Notice that
$$ \{T_1 = 2n+1\} = \{S_1 \leq 0, \cdots, S_{2n-1} \leq 0, S_{2n} = 0, S_{2n+1} = 1\}. $$
Using one of the equivalent characterization of Catalan number, we can explicitly compute the probability of this event as
$$ \Bbb{P}(T_1 = 2n+1) = \frac{1}{2^{2n+1}}C_n = \frac{1}{2^{2n+1}(n+1)} \binom{2n}{n}. $$
From this, we explicitly compute the probability generating function of $T_1$ by
$$ |z| < 1 \quad \Rightarrow \quad \mathbb{E}[z^{T_1}] = \sum_{n=0}^{\infty} z^n \mathbb{P}(T_1 = n) = \sum_{n=0}^{\infty} \frac{C_n}{2^{2n+1}} z^{2n+1} = \frac{1-\sqrt{1-z^2}}{z}. $$
Letting $z \to 1^-$ shows that, by the monotone convergence theorem,
$$ \mathbb{P}(T_1 < \infty) = \lim_{z \to 1^-} \mathbb{E}[z^{T_1}] = \lim_{z \to 1^-} \frac{1-\sqrt{1-z^2}}{z} = 1. $$
Therefore $\Bbb{P}(T_1 = \infty) = 0$.
Addendum. Using this, we can also show that
$$ \mathbb{E}[T_1 z^{T_1}] = \frac{1-\sqrt{1-z^2}}{z\sqrt{1-z^2}}, $$
and so, $T_1$ infinite expectation $\Bbb{E}[T_1] = \infty$.
Best Answer
Your proof is correct: the expected value of this hitting time is infinite.
To add, $S_n$ is a Markov Chain (symmetric random walk) on the integers. It is null recurrent. The proof you just gave verifies it cannot be positively recurrent. -- Lost1