Uniform normal filter on $\kappa$ is $\kappa$-complete and contains the club filter. The question reads:
Edit: The definition for uniform filter I was given is as follows: A filter $\mathcal{F}$ on $\kappa$ is uniform iff $\mathcal{F}$ contains all co-bounded sets in $\kappa$ iff $\mathcal{F}$ contains all intervals $(\alpha, \kappa)$ for $\alpha < \kappa$.  This may not be standard, it seems uniformity usually only applies to ultrafilters.
"Work in $\sf ZF$.  Assume $\kappa > \omega$ is regular and $\mathcal{F}$ is a uniform normal filter on $\kappa$.  


*

*Prove that $\mathcal{F}$ is $\kappa$-complete.

*Prove that $\mathcal{F}^{\text{club}}_\kappa \subseteq \mathcal{F}$


Hint: For 1 for an ordinal $\lambda < \kappa$ and a sequence of sets $\langle A_\xi \mid \xi < \lambda \rangle$ of sets in $\mathcal{F}$, extend this sequence to a sequence of length $\kappa$ whose diagonal intersection differs from $\bigcap_{\xi < \lambda} A_\xi$ by a non-stationary set.  For 2 given a club set $C \subseteq \kappa$, try to express $C$ as a diagonal intersection of intervals of the form $(\alpha, \kappa)$.  Alternatively, using choice, use pressing down."
So I believe, I found the extended sequence for 1, but I'm not sure why having the diagonal intersection differ by a non-stationary set helps.  For 2, I couldn't figure out the intervals.  I tried enumerating the club $C$ in increasing order, and tried to create the intervals that way, but I can't seem to get anywhere.   I believe, that it is sufficient to just consider club $C$ since any $X \in \mathcal{F}^{\text{club}}_\kappa$ contains a club so then by the upward-closure of $\mathcal{F}$ we will get $X \in \mathcal{F}$.  I'm not sure how to do the pressing down argument, but I would love to see that.
My (partial) solution to 1.
Let $\lambda < \kappa$ and $\langle A_\xi \mid \xi < \lambda \rangle \subseteq \mathcal{F}$.  Let $A := \bigcap_{\xi <\kappa} A_\xi$.  Define a new sequence $\langle A'_\xi \mid \xi < \kappa \rangle$ by
$$A'_\xi = \left\{
        \begin{array}{ll}
            A_\xi & \quad \xi < \lambda \\
            (\delta, \kappa) & \quad \lambda \leq \delta < \kappa
        \end{array}
    \right.$$
Since $\mathcal{F}$ is uniform, $(\delta, \kappa) \in \mathcal{F}$ for all $\delta < \kappa$ so that $\langle A'_\xi \mid \xi < \kappa \rangle$ is a sequence in $\mathcal{F}$.  By normality, $A' := \bigtriangleup_{\xi < \kappa} A'_\xi \in \mathcal{F}$.  Then $(A - A') \cup (A' - A)  \subseteq [0, \lambda]$ which is bounded, so it is non-stationary.
 A: We have that $A\triangle A'$ is bounded; as you noted, thus by your definition of uniform filter we have that $(A\triangle A')^c\in\mathcal F$, but $(A\triangle A')^c=(A^c\cap A'^c)\cup(A'\cap A),$ hence as $A'\in\mathcal F$ we obtain, intersecting, that $(A'\cap A)\in\mathcal F$, which implies $A\in\mathcal F$.
For $(2)$, let $X_{\alpha}=(\alpha+1,\kappa),$ for each $\alpha<\kappa$, then each $X_{\alpha}\in\mathcal F$, and clearly $C_0=\triangle_{\alpha<\kappa}X_{\alpha}\in\mathcal F$, and $C_0$ is the set of all limit ordinals $<\kappa$. 
Now let $C$ be a closed unbounded set. Let $\{a_{\xi}:\xi<\kappa\}$ be the increasing continuous enumeration of $C$. For each $\alpha<\kappa$ let $Y_{\alpha}=(a_{\alpha},\kappa)$. Let $\xi\in C_0\cap\triangle_{\alpha<\kappa}Y_{\alpha},$ then $a_{\alpha}<\xi$ for all $\alpha<\xi$, thus if $\beta=\lim_{\alpha\to\xi}a_{\alpha},$ then $\beta=a_{\xi}$ and $\beta\leq\xi$, thus $\xi\in C$. Hence $C\supseteq C_0\cap\triangle_{\alpha<\kappa}Y_{\alpha}$, but $\triangle_{\alpha<\kappa}Y_{\alpha}\in\mathcal F$, therefore $C\in\mathcal F$.
