[proofplan]
The proof shows that every singleton is outside $U$, hence every singleton complement is inside $U$ by the ultrafilter property. Since $A$ has cardinality strictly less than $\kappa$, $\kappa$-completeness permits intersecting the family of complements $\kappa \setminus \{\beta\}$ for $\beta \in A$. This intersection is exactly $\kappa \setminus A$, so $A$ cannot also lie in the filter. The bounded-set and tail conclusions follow because $\kappa$ is regarded as an initial ordinal, so every ordinal $\alpha < \kappa$ has cardinality less than $\kappa$.
[/proofplan]
[step:Put the complement of each singleton into the ultrafilter]
Let $\beta \in \kappa$. Since $U$ is nonprincipal, the singleton $\{\beta\}$ does not belong to $U$. Since $U$ is an ultrafilter on $\kappa$, exactly one of a subset of $\kappa$ and its complement belongs to $U$. Therefore
\begin{align*}
\kappa \setminus \{\beta\} \in U.
\end{align*}
[guided]
Fix an element $\beta \in \kappa$. The first use of nonprincipality is to rule out the possibility that $U$ concentrates at the point $\beta$: by definition of nonprincipal ultrafilter, no singleton subset of $\kappa$ belongs to $U$, so
\begin{align*}
\{\beta\} \notin U.
\end{align*}
Now we use the ultrafilter property. For every subset $E \subset \kappa$, an ultrafilter contains exactly one of $E$ and $\kappa \setminus E$. Applying this with $E = \{\beta\}$ gives
\begin{align*}
\kappa \setminus \{\beta\} \in U.
\end{align*}
This is the pointwise input for the later $\kappa$-complete intersection: every individual point of $A$ can be removed while staying inside the ultrafilter.
[/guided]
[/step]
[step:Intersect fewer than $\kappa$ singleton complements]
Let $A \subset \kappa$ satisfy $|A| < \kappa$. If $A = \varnothing$, then $\kappa \setminus A = \kappa \in U$ because every filter on $\kappa$ contains the whole underlying set. Suppose now that $A \neq \varnothing$.
For each $\beta \in A$, the previous step gives $\kappa \setminus \{\beta\} \in U$. The family
\begin{align*}
\mathcal{F}_A := \{\kappa \setminus \{\beta\} : \beta \in A\}
\end{align*}
has cardinality at most $|A|$, hence has cardinality strictly less than $\kappa$. Since $U$ is $\kappa$-complete, the intersection of this fewer-than-$\kappa$ family of members of $U$ also belongs to $U$:
\begin{align*}
\bigcap_{\beta \in A}(\kappa \setminus \{\beta\}) \in U.
\end{align*}
By the elementary set identity
\begin{align*}
\bigcap_{\beta \in A}(\kappa \setminus \{\beta\}) = \kappa \setminus A,
\end{align*}
we obtain
\begin{align*}
\kappa \setminus A \in U.
\end{align*}
[/step]
[step:Exclude the small set itself from the ultrafilter]
Assume, toward a contradiction, that $A \in U$. From the previous step, $\kappa \setminus A \in U$. Since $U$ is a filter, it is closed under finite intersections, so
\begin{align*}
A \cap (\kappa \setminus A) \in U.
\end{align*}
But
\begin{align*}
A \cap (\kappa \setminus A) = \varnothing,
\end{align*}
and no filter contains $\varnothing$. This contradiction proves $A \notin U$. Thus every $A \subset \kappa$ with $|A| < \kappa$ satisfies $A \notin U$ and $\kappa \setminus A \in U$.
[/step]
[step:Apply the small set result to bounded subsets and tails]
Let $B \subset \kappa$ be bounded in $\kappa$. By definition of boundedness in the ordinal $\kappa$, there exists an ordinal $\alpha < \kappa$ such that $B \subset \alpha$. Since $\kappa$ is regarded as an initial ordinal, every ordinal $\alpha < \kappa$ has cardinality strictly less than $\kappa$. Therefore
\begin{align*}
|B| \leq |\alpha| < \kappa.
\end{align*}
Applying the result already proved with $A = B$ gives
\begin{align*}
\kappa \setminus B \in U.
\end{align*}
Finally, fix $\alpha < \kappa$. The set $\alpha = \{\beta \in \kappa : \beta < \alpha\}$ is bounded in $\kappa$, so the preceding paragraph applied to $B = \alpha$ gives
\begin{align*}
\kappa \setminus \alpha \in U.
\end{align*}
This proves the bounded-set conclusion and the tail conclusion.
[/step]
