$(a_n)_n\subset [c,d]$ with limit point $h$ and $f: [c,d] \to \mathbb{R}$ continuous $\implies$ $f(h)$ is a limit point of $(f(a_n))_n$ 
$(a_n)_n\subset [c,d]$ with limit point $h$ and $f: [c,d] \to
 \mathbb{R}$ continuous $\implies$ $f(h)$ is a limit point of
  $(f(a_n))_n$

My attempt:
Let $f: [c,d]\to \mathbb{R}$ be continuous. Since $(a_n)_n\subset [c,d]$ has a limit point $h$, there is $(a_{n_k})_k$ such that $a_{n_k}\to h$. Since $f$ is continuous, for every $\varepsilon >0$, there is $\delta >0$ such that for all $x,h\in [c,d]$: $$|x-h|<\delta \implies |f(x)-f(h)|<\varepsilon$$. Particularly, for every $k\in \mathbb{N}$, there is a $x_{n_k}\in [c,d]$ such that $|x_{n_k}-h|<\delta \implies |f(x_{n_k})-f(h)|<\varepsilon$.

The last sentence is not correct. Since $a_{n_k}\to h$, there is $K$ s.t. $|a_{n_k}-h|<\delta $ for all $k\geq K$. Therefore, $|f(a_{n_k})-f(h)|<\varepsilon $ when $k\geq K$, what prove the claim.
