Let $\left({E, \preceq}\right)$ be a poset, and let $K \subseteq E$ be a maximal chain in $E$ such that $K$ has an upper bound in $E$. Then $K$ has at least one element.
Proof: $K$ is a maximal chain. That is, for every element $x \in E \setminus K$ there is a $y \in K$ such that $(x,y) \notin \, \preceq$ and $(y,x) \notin \, \preceq$. This means that if an $x \in E$ belongs to $E \setminus K$, we surely have an element of $K$ that isn't comparable with this $x$ (and so no element of the chain is comparable with $x$, because of transitivity). Now let $m \in E$ be an upper bound for $K$; we can compare $m$ with any element the chain. But this is the same as saying that $m \notin E \setminus K$ and so $m \in K$. $\blacksquare$
In other words, if $m$ is an upper bound for $K$, it must belongs to $K$ because it has to be comparable with all elements of $K$ itself, and we said that $K$ is maximal (if we are able to compare $m \notin K$ with every element of the chain, it's reasonable to "expand" it including $m$, in contrast with the hypothesis that $K$ is maximal).
Is it a correct proof?
Yes, you have proved that an upper bound of a maximal chain is an element of the chain (and thus a maximum of that chain). So it has to be non-empty.
Another way of looking at this is that if the empty set is a maximal chain, then the poset is empty, and there is no upper bounds of anything.