Random walk to stay in an interval forever

Yes. Indeed, if $s = \sum_{i \geq 1} t_i^2 <1$, then $$ \mathbb{P}[ \ \ \forall n, \sum_{i=1}^n X_i \in [-1,1] \ \ ] \geq 1-s > 0. $$ To see this, note that $M_n = |\sum_{i=1}^n X_i|$ is a nonnegative submartingale, so that Doob's martingale inequality yields $$ \mathbb{P}[ \max_{1 \leq j \leq n} M_j > 1 ] \leq \mathbb{E}[M_n^2] = \sum_{i=1}^n t_i^2 \leq s. $$ Letting $n$ tends to infinity, one gets $$ \mathbb{P}[ \sup_{j \geq 1} M_j > 1 ] \leq s, $$ hence the result by taking the complement.

Remark : In the other direction, one can show that if $$ \mathbb{P}[ \ \ \forall n, \sum_{i=1}^n X_i \in [-1,1] \ \ ] \geq c > 0, $$ then $$ s = \sum_{i \geq 1} t_i^2 \leq \frac{14}{c^2}. $$ One first note that one must have $|t_i| \leq 2$ for all $i$. One has $|\cos(2 \pi \xi)| \leq e^{-2 \pi^2 \xi^2 }$ whenever $|\xi| \leq 2 \delta$, with $\delta = 0.14$, hence $$ |\mathbb{E}[e^{-2i \pi \xi \sum_{j=1}^n X_j} ]| = \prod_{j=1}^n |\cos(2 \pi \xi t_j)| \leq e^{-2 \pi^2 \xi^2 \sum_{i=1}^n t_i^2} $$ whenever $|\xi| \leq \delta$. Let $\chi$ be the Beurling-Selberg majorant of $\mathbb{1}_{[-1,1]}$ with parameter $\delta$ : one has $\chi \geq \mathbb{1}_{[-1,1]}$, $||\chi||_{L^1}= 2 + \delta^{-1}$, and its Fourier transform $\hat{\chi}$ is supported on $[-\delta,\delta]$. In particular : $$ c \leq \mathbb{E}[\chi(\sum_{i=1}^n X_i)] = \int_{\mathbb{R}} \hat{\chi}(\xi) \mathbb{E}[e^{-2i \pi \xi \sum_{j=1}^n X_j} ] d \xi \leq (2 + \delta^{-1}) \int_{\mathbb{R}} e^{-2 \pi^2 \xi^2 \sum_{i=1}^n t_i^2} d \xi, $$ and thus $$ c \leq \frac{2 + \delta^{-1}}{\sqrt{2 \pi \sum_{i=1}^n t_i^2}}. $$ This implies $$ \sum_{i=1}^n t_i^2 \leq \frac{(2 + \delta^{-1})^2}{2 \pi c^2} \leq \frac{14}{c^2}, $$ hence the result.

The crucial requirement is that $\sum_{i=0}^\infty t_i^2 < \infty$. See Kolmogorov's two-series theorem and also the more general Kolmogorov's three-series theorem.