A set $A$ is countably compact iff every countable subset (or sequence of distinct points, if you prefer) of $A$ has a cluster point in $A$.
The alternative formulation is
$(\ast$) For all $A \subseteq X$: if $A$ is not closed there exist $a_n \in A$ such that $A$ contains no cluster point for $(a_n)$.
The proposed equivalence is simply a matter of contrapositive reasoning:
Suppose $X$ is $C$-closed. Let $A$ be non-closed. This means that $A$ is not countably compact. So there is a sequence $x_n$ from $A$ without a cluster point in $A$. Hence $X$ obeys ($\ast$). For the reverse, suppose $X$ obeys the property $(\ast)$. Then $X$ is $C$-closed: let $A$ be countably compact. Then $A$ is closed. For suppose it were not closed, then by the assumption $(\ast)$ on $X$ we'd have a sequence $a_n$ from $A$ without a cluster point in $A$. This contradicts $A$ being countably compact. So $A$ is closed and $X$ is $C$-closed.
The example $X= \omega_1+1$ is no contradiction. It is indeed not $C$-closed. The negation of $C$-closed in the alternative formulation ($\ast$) is:
There exists a subset $A$ of $X$ that is not closed, but such that all sequences from $A$ have a cluster point in $A$.
And indeed $A = \omega_1$ is as required. So $X$ is also not $C$-compact in the other definition ($\ast$).
I think it's mostly a matter of taking the negation of the alternative definition in the right way.