Prove that if $b_n$ is a subsequence of $a_n$ and $c_n$ is a subsequence of $b_n$, then $c_n$ is a subsequence of $a_n$. Let $(a_{n})_{n=0}^{\infty}$, $(b_{n})_{n=0}^{\infty}$ and $(c_{n})_{n=m}^{\infty}$ be sequences of real numbers. Then $(a_{n})_{n=0}^{\infty}$ is a subsequence of $(a_{n})_{n=0}^{\infty}$. Furthermore, if $(b_{n})_{n=0}^{\infty}$ is a subsequence of $(a_{n})_{n=0}^{\infty}$, and $(c_{n})_{n=0}^{\infty}$ is a subsequence of $(b_{n})_{n=0}^{\infty}$, then $(c_{n})_{n=0}^{\infty}$ is a subsequence of $(a_{n})_{n=0}^{\infty}$.
My solution
We say that a sequence of real numbers $(x_{n})_{n=0}^{\infty}$ is a subsequence of $(y_{n})_{n=0}^{\infty}$ if there exists a strictly increasing function $f:\textbf{N}\to\textbf{N}$ such that $x_{n} = y_{f(n)}$.
Based on this definition, we conclude that $a_{n}$ is a subsequence of itself: it suffices to choose $f(n) = n$.
On the other hand, if $b_{n}$ is a subsequence of $a_{n}$ and $c_{n}$ is a subsequence of $b_{n}$, then exist functions $f:\textbf{N}\to\textbf{N}$ and $g:\textbf{N}\to\textbf{N}$ such that $b_{n} = a_{f(n)}$ and $c_{n} = b_{g(n)}$.
Consequently, $c_{n} = a_{f(g(n))}$, where $f\circ g$ is strictly increasing because it is a composition of two strictly increasing functions.
Could someone please tell me if I am missing any formal step? Any comment is appreciated!

No steps missing, but there is an error.   If $b_n = a_{f(n)}$, then 
$$  c_n = b_{g(n)} = a_{f(g(n))}  \text{.}  $$
You have the composition reversed.
