Is closed subspace (open subspace) of a sequential space sequential? 
A topological space is called a sequential space if a set $A ⊂ X$ is
  closed if and only if together with any sequence it contains all its
  limits.

Is every open subspace of a sequential space, still a sequential space? or every closed subspace of a sequential space, still a sequential space?
 A: I will freely use these definitions.
Let $X$ be a sequential space, and suppose that $F$ is a closed subspace of $X$. Let $A$ be a sequentially closed subset of $F$. Suppose that $\sigma$ is a sequence in $A$ converging to some point $x\in X$; $\sigma$ is a sequence in $F$, and $F$ is closed in $X$, so $x\in F$, and since $A$ is sequentially closed in $F$, it follows that $x\in A$. Thus, $A$ is sequentially closed in $X$ as well as in $F$, and since $X$ is sequential, $A$ is closed in $X$. But $A\subseteq F$, so $A=A\cap F$, and therefore $A$ is closed in $F$. This shows that every sequentially closed subset of $F$ is closed in $F$ and hence that $F$ is sequentially closed.
Now suppose that $U$ is an open subspace of $X$, and let $V$ be a sequentially open subset of $U$. Let $\sigma=\langle x_n:n\in\omega\rangle$ be any sequence in $X$ converging to some $x\in V$; $x\in U$, which is open in $X$, so $\sigma$ is eventually in $U$, i.e., there is an $m\in\omega$ such that $x_n\in U$ for each $n\ge m$. Then $\sigma'=\langle x_n:n\ge m\rangle$ is a sequence in $U$ converging to $x\in V$, and $V$ is sequentially open in $U$, so $\sigma'$ is eventually in $V$: there is an $m'\ge m$ such that $x_n\in V$ for each $n\ge m'$. Thus, $\sigma$ is eventually in $V$, and $V$ is therefore sequentially open in $X$ as well as in $U$. Finally, $X$ is sequential, so $V$ is actually open in $X$. Finally, $V=V\cap U$, so $V$ is open in $U$, and it follows that $U$ is sequential.
