Let $G$ be a profinite group, and $U \leq_O G$ (i.e., $U$ is an open subgroup of $G$) be strongly complete. Then $G$ is strongly complete.
This problem is actually stated as its contrapositive (if $G$ is not strongly complete, then no open subgroup of it is), but the only reasonable approach I can see to that would be a contradiction argument which basically runs exactly the same way as this one. (Strongly complete means "every finite-index subgroup is open" or equivalently, "...is closed")
It's enough to show that $V \triangleleft_f G$ (subscript "f" means "finite index") implies $V \triangleleft_O G$. In this case, certainly $U \cap V \leq_f U$, so $U \cap V \leq_O U$ since $U$ is strongly complete. But then $U \cap V \leq_O U \leq_O G$ implies $U \cap V \leq_O G$. This seems to be halfway there, I just need to somehow show that $U^c \cap V$ is an open set since $V = (U \cap V) \cup (U^c \cap V)$, but (if this is the right strategy) here I get stuck. Sure, $U \cap V$ is also closed, so $(U \cap V)^c$ is open, but that doesn't seem like quite enough to work with.
Let $V$ be a finite index subgroup of $G$. Then as $g$ varies, $gU$ is an open cover of $G$. Thus to show that $V$ is open in $G$, it suffices to show that $gU\cap V$ is open for all $g$. Now $gU\cap V$ is open if and only if $U\cap g^{-1} V$ is open. But if $U\cap g^{-1}V$ contains some element $h$, then $U\cap g^{-1}V = h(U\cap V)$, i.e. it is a coset of $U\cap V$ in $U$. Then if $U\cap V$ has finite index in $U$, we'll have that $h(U\cap V)$ is open, hence that $gU\cap V$ is open.
Hence to show that $gU\cap V$ is open for all $g$, it suffices to show that $U\cap V$ has finite index in $U$. However, this is clear, since if $h(U\cap V)\ne h'(U\cap V)$ for $h,h'\in U$, but $hV = h'V$, then $hv=h'v'$ for $v,v'\in V$, so $h'^{-1}h=v'v^{-1}\in U\cap V$. This would contradict the assumption that $h(U\cap V)\ne h'(U\cap V)$. Hence there are at least as many cosets of $V$ in $G$ as there are of $U\cap V$ in $U$. Hence $U\cap V$ has finite index in $U$ as desired.
Thus $V$ must be open in $G$ if it has finite index in $G$.