Theorem: Let $(a_n)_{n\in \mathbb{N}},(b_n)_{n\in \mathbb{N}} \subset \overline{\mathbb{R}}$ be two sequences, and suppose that $\exists \nu \in \mathbb{N}: a_n \le b_n \forall n \ge \nu$. Then: $$ \liminf_n a_n \le \liminf_n b_n $$
My proof attempt
$$a_n \le b_n \quad\forall n \ge \nu \Longrightarrow \inf_{n \ge k}a_n \le a_k \le b_k \quad\forall k \ge \nu \Longrightarrow \inf_{n \ge k}a_n \le \inf_{n \ge k}b_n \quad \forall k \ge \nu$$
$c_k := \inf_{n \ge k}a_n$ and $g_k := \inf_{n \ge k}b_n$ then:
$$ c_k \le g_k \quad \forall k \ge \nu \Longrightarrow \sup_{k\ge h}g_k \ge g_h \ge c_h \quad \forall h \ge \nu \Longrightarrow \sup_{k\ge h}g_k \ge \sup_{k\ge h}c_k \quad \forall h \ge \nu $$
Now $\liminf_k b_k= \sup_{k \in \mathbb{N}}g_k$ but I have $\sup_{k\ge h \ge \nu}g_k$.
However $$\sup_{k \in \mathbb{N}}g_k \ge \sup_{k\ge h \ge \nu}g_k$$ because on the LHS we are taking the supremum on a set whose cardinality is greater or equal (if $\nu=1$) to the one of the set on the RHS.
Therefore:
$$\liminf_k b_k=\sup_{k \in \mathbb{N}}g_k \ge \sup_{k\ge h \ge \nu}g_k \ge \sup_{k\ge h}c_k $$
Problem
But $\sup_{k\ge h}c_k \le \sup_{k \in \mathbb{N}}c_k = \liminf_k a_k$ and then:
$$\liminf_k b_k=\sup_{k \in \mathbb{N}}g_k \ge \sup_{k\ge h \ge \nu}g_k \ge \sup_{k\ge h}c_k \le \sup_{k \in \mathbb{N}}c_k = \liminf_k a_k$$
which clearly doesn't prove the above theorem.
Where am I doing wrong?
Edit: For this proof attempt I followed the line of reasoning of the answers provided here
Your proof is correct except that $g_k, c_k$ are an increasing sequence w.r.t $k$ so $\sup_{k \geq 0} g_k = \sup_{k \geq h \geq v} g_k$ and $\sup_{k \geq 0} c_k = \sup_{k \geq h \geq v} c_k$.