Let $x_n$ be a convergent sequence $x_n \to x$ in a topological space $X$ and $F:X\longrightarrow \mathbb R$ any function (not necessarily continuous). If there exist a subsequence $x_{n_k}$ such that
$$F(x) \leq \liminf_k F(x_{n_k})$$
then can we conclude that
$$F(x) \leq \liminf_n F(x_n) \,\,\, ?$$
Let $X = \mathbb{R}$. Let $x_n$ be a sequence converging to $0$ such that $x_n$ is rational if $n$ is odd, and irrational if $n$ is even. Let $F = \mathbb{1}_{\Bbb{Q}}$ be the rational indicator function. Finally, let $(x_{n_k})$ be the subsequence of all odd indices, so $x_{n_k} \in \Bbb{Q}$ for all $k$. Then: $$ \liminf_k F(x_{n_k}) = \liminf_k 1 = 1 \geq F(x) $$ But: $$ \liminf_n F(x_n) = 0 \not\geq F(x) $$ as $F(0) = 1$.