Does $g(v_n) longrightarrow g(0)$ for all $v_n text{s.t.} ||v_{n+1}|| leq ||v_n||$ imply $g$ continuos at $0$?

Assume that $v_n rightarrow 0$ and $g(v_n) nrightarrow g(0)$. Then there exists a subsequence $v_{k_n}$ s.t. $||g(v_{k_n}) - g(0)|| ge varepsilon$ for some $varepsilon > 0$. Let $u_n$ denote the sequence $v_{k_n}$. There exists a subsequence $u_{m_n}$ s.t. $||u_{m_{n+1}}||le ||u_{m_n}||$. Then $g(u_{m_n}) rightarrow g(0)$. But it is already known that $||g(u_{m_n}) - g(0)|| ge varepsilon$. This contradiction shows that $g(v_n) rightarrow g(0)$ for all $v_n rightarrow 0$. Thus, $g$ is continuous at $0$.

Take an arbitrary subsequence ${g(v_{n_k})}_k$ and then choose a subsequence ${v_{n_{k_l}}}_l$ such that $| v_{n_{k_{l+1}}} | leq | v_{n_{k_{l}}} |$. Using the assumptions we can deduce $g(v_{n_{k_l}}) to g(0)$. Why does this imply $g(v_n) to g(0)$ already?