Suppose that $$ \liminf_{k \to \infty} \|\nabla f(x_k)\| = 0 $$ and the sequence $\{x_k\}$ is bounded. Is it true that all accumulation points of $\{x_k\}$ are stationary points?
Can you find a subsequence $\{x_{k_m}\}$ such that $x_{k_m}$ converges to a stationary point?
Let's assume that $f \in C^1(\mathbb R^n)$.
Obviously, not all accumulation points are stationary solutions, since the liminf gives us only information about one subsequence of $\{\nabla f(x_k)\}$.
Counterexample: just consider the sequence $x_k = 0, x_{k+1}=1$ for $k \in \mathbb N$ even and $f(x)=x^2$.
To your second question: This should be true if $f$ is continuously differentiable. Let $\{y_k\} \subset \mathbb R^n$ with $y_k \to y$ and $$ \lim_{k \to \infty} \vert \nabla f(y_k) \vert = 0. $$ Since $\nabla f$ is continuous, we have $$ \vert \nabla f(y) \vert = \lim_{k \to \infty} \vert \nabla f(y_k) \vert = 0. $$ Therefore, $y$ is a stationary solution of $f$.