Let $\mathfrak U$ be an algebra (i.e. a set, called universe, together with several $n$-ary operations) in the sense of universal algebra. Denote by $\operatorname{Con} \mathfrak U$ the set of all congruence relations on $\mathfrak U$. A congruence $\theta$ on $\mathfrak U$ is said to be fully invariant if the following compatibility condition is valid for any endomorphism $\phi$ of $\mathfrak U$ and $a,b \in A$: $$ a \theta b \Rightarrow \phi a \theta \phi b. $$ The set of all fully invariant congruences on $\mathfrak U$ will be represented by $\operatorname{Con}_{fi} \mathfrak U$. Both $\operatorname{Con} \mathfrak U$ and $\operatorname{Con}_{fi} \mathfrak U$, ordered by the relation of inclusion, are lattices. In fact, the intersection of (fully invariant) congruences on $\mathfrak U$ is still a (fully invariant) congruence on $\mathfrak U$.
Exercise: Show that $\operatorname{Con}_{fi} \mathfrak U$ is a complete sublattice of the lattice $\operatorname{Con} \mathfrak U$ of all congruences on $\mathfrak U$, i.e. for any nonempty subset $C$ of $\operatorname{Con}_{fi} \mathfrak U$ the infima (respectively, the supremum) of $C$ in the two lattices coincide.
My solution (for infimum), let $C \subseteq \operatorname{Con}_{fi} \mathfrak U \subseteq \operatorname{Con} \mathfrak U$, then $$ \theta = \bigcap_{\theta' \in C} \theta' $$ is a congruence (so also in $\operatorname{Con} \mathfrak U$), which is obviously fully invariant (so also in $\operatorname{Con}_{fi} \mathfrak U$). And because the intersection of sets is unique, the two infima (which are defined in both cases as the intersection) coincide.
This is my proof, but think this is quite trivial and simple, so my question does I have overlooked something or is my proof valid?
It appears that your proof (that the infima coincide) is valid. However, the difficulty in showing that a set is a complete sublattice usually occurs when trying to show that the suprema coincide.
For example, in this problem, you would need to show that the complete join of fully invariant congruences (as computed in $\mathrm{Con}\,\mathfrak{A}$) is a fully invariant congruence. In other words, you would have to show the complete join as computed in $\mathrm{Con}\,\mathfrak{A}$ is the same as the complete join as computed in $\mathrm{Con}_{fi}\,\mathfrak{A}$.