An unbounded subset of $\kappa$ is some $A\subseteq\kappa$ such that for all $\beta<\kappa$, there is some $\alpha\in A$ such that $\beta<\alpha$.
A closed subset of $\kappa$ is some $A\subseteq\kappa$ such that for all $0<\alpha<\kappa$, if $\sup(A\cap\alpha)=\alpha,$ then $\alpha\in A$.
For the first one, do you know the result that all limit ordinals are of the form $\omega\cdot\alpha$ for some non-zero $\alpha$? In particular, you'll want to show that the limit ordinals less than $\kappa$ are those of the form $\omega\cdot\alpha$ for $0<\alpha<\kappa$. Use continuity of ordinal multiplication to show that this set is closed, and the fact that $\kappa$ is a cardinal to show that it's unbounded.
The second one is just a special case of the first one. (Why?)
The third one won't be too tricky. Do you know the recursive definition of the alephs? (Also, don't forget that the natural numbers are cardinals, too.)
Closed means that any limit of elements of $A$, are also in $A$, so let's take increasing a sequence $(α_i ∈ A)_{i∈ω}$, and we will show that $\lim α_i=α∈ A$ (I assume monotocity, so $\lim$ here is the same as $\sup$).
We know that each $α_i$ is a limit of some increasing sequence $(β_i^j ∈ X)_{j\in\omega}$, define $f:ω→ω$ to be $f(i)=\min(j\in\omega\mid β_{i+1}^j>α_i)$ (note, it is well defined because $α_i=\lim_j β_i^j<\lim_j β_{i+1}^j=α_{i+1}$, also note that $f$ is not necessarily increasing).
I claim that $\lim_i β_{i+1}^{f(i)}=α$, indeed $β_{i+1}^{f(i)}<α_{i+1}<α$ for all $i$, so $\lim β_{i+1}^{f(i)} ≤α$, but if $x<α$, there exists $i$ such that $x<α_i<α$, and so $x<α_i<β_{i+1}^{f(i)}<α$, hence $\lim_i β_{i+1}^{f(i)}=α$.
So, given any limit point $α$ of $A$, we have a (countable) sequence from $X$, whose limit is $α$, hence $α∈A$
Best Answer
Fix $\xi\in \kappa$, since $A$ is unbounded there is a $\alpha_0\in A$ so that $\xi<\alpha_0$. Now, construct recursively a strictly increasing sequence $\langle \alpha_n: n\in \omega\rangle$. Let $\alpha=\sup\{\alpha_n: n\in \omega\}.$ Since $\kappa$ is regular and uncountable, we have $\alpha<\kappa.$ It is also easy to see that $\sup(A\cap\alpha)=\alpha$.