Let $(X_n)_{n\in\mathbb N}$ be a real-valued process on a probability space $(\Omega,\mathcal A,\operatorname P)$ with $$\liminf_{n\to\infty}X_n=-\infty\;\;\;\text{almost surely}.\tag1$$ Are we able to conclude $$\liminf_{n\to\infty}\operatorname E\left[1\wedge e^{X_n}\right]=0?\tag2$$
We may note that $\mathbb R\ni x\mapsto1\wedge x$ is Lipschitz continuous. Now, it looks like some kind of dominated convergence, but with $\liminf$ instead of $\lim$. Another approach might be to note that $$\operatorname E\left[1\wedge e^{X_n}\right]=\int_0^1\operatorname P\left[e^{X_n}\ge t\right]\:{\rm d}t\tag3.$$
EDIT: Maybe the claim is easier to prove in greater generality: Let $f:\mathbb R\to\mathbb R$ be bounded and Lipschitz continuous with $$f(x)\xrightarrow{x\to-\infty}0.$$ By $(1)$, there is an increasing $(n_k)_{k\in\mathbb N}\subseteq\mathbb N$ with $$X_{n_k}\xrightarrow{k\to\infty}-\infty\;\;\;\text{almost surely}\tag4$$ and hence $$f\left(X_{n_k}\right)\xrightarrow{k\to\infty}0\;\;\;\text{almost surely}.\tag5$$ Since $f$ is bounded, we obtain $$\operatorname E\left[f\left(X_{n_k}\right)\right]\xrightarrow{k\to\infty}0\tag6$$ by dominated convergence. Is this enough to conclude?
Here is a counter-example: Let $Z_n$ be a sequence of i.i.d. random variables with $Z_n\sim\operatorname{Bernoulli}(\frac{1}{2})$, and set $X_n = -n Z_n$. Then $\liminf_{n\to\infty} Z_n = -\infty$ almost surely, but
$$\mathbb{E}[1 \wedge e^{X_n}] \geq \frac{1}{2} \quad \text{for all} \quad n \geq 1,$$
since $\mathbb{P}(X_n = 0) = \frac{1}{2}$. By changing the parameter of the Bernoulli distribution, we can come up with examples with the lower bound as close to 1 as we want.
The issue in your argument is that the subsequence realizing $f(X_{n_k}) \to -\infty$ need not be deterministic in general. And even if $N_k$ is a measurable choice of subsequence so that $X_{N_k} \to -\infty$ almost surely, we have no good way of relating this to the expectation of $f(X_n)$ simply because $N_k$ is not deterministic.