In Schaefer's book Topological Vector Spaces, if $\Bbb K=\Bbb R\hbox{ or }\Bbb C$, and if $E\in\mathsf{Vec}_{\Bbb K}$, then if we set $\mathscr{T}_{\mathrm{LC}}$ to be the family of locally convex topologies on $E$, then we seem to be implicitly assuming that $\mathscr{T}_{\mathrm{LC}}$ is a complete lattice.
Now, I have been able to prove that $\mathscr{T}_{\mathrm{LC}}$ is a lattice, and that if $(\tau_\alpha)_{\alpha\in A}$ is an ascending (i.e. increasingly fine) chain, then it has a $\sup$. However, if $(\tau_\alpha)_{\alpha\in A}$ is a descending (i.e. increasingly coarse) chain, then I fail to see why it has an $\inf$. My difficulty is the following:
If we let $\tau$ be the topology consisting of unions of elements of $\bigcap_{\alpha\in A}\tau_\alpha$, then $\tau$ is clearly a translation-invariant topology, so I wish to show that if $0\in U\in\tau_\alpha$ for all $\alpha$, then there exists some circled radial convex (c.r.c.) neighborhood $V\in\tau$ such that $0\in V\subset U$, which would be enough to show that $\tau$ is a locally convex topology. But, all I know is that for all $\alpha\in A$, there exists some c.r.c. $V_{\alpha}\in\tau_\alpha$ such that $0\in V_\alpha\subset U$, and so I'm tempted to assume that $A=\mathbb{N}$ and set $$V=\sum_{\alpha=1}^\infty 2^{-\alpha}V_\alpha\in\tau$$ but this only works for countable $A$, and in any case, without knowing that $U$ is convex (which is pretty close to what we want to prove in any case) we do not know that $V\subset U$.
So I'm presuming my approach is the wrong one, but then how would one go about proving this?
Note that for a vector space $E$ with translation-invariant topology $\tau$, $(E,\tau)$ is a locally convex topology iff there exists a $0$-neighborhood base $\mathcal{B}$ consisting of circled radial convex sets, such that $\lambda\in\Bbb K\setminus\{0\}$ and $U\in\mathcal{B}$ implies that $\lambda U\in\mathcal{B}$.
To prove a poset $P$ is a complete lattice, you only need to prove that it has arbitrary joins, since then it follows formally that $P$ has arbitrary meets as well. Explicitly, let $S\subseteq P$ be any subset and let $T$ be the set of lower bounds of $S$. Since $P$ has arbitrary joins, the set $T$ has a join $p\in P$. Then $p$ is itself a lower bound of $S$: every element of $S$ is an upper bound of $T$, and is thus greater than or equal to $p$. It follows that $p$ is the meet of $S$.
So in your case, having shown that $\mathscr{T}_{\mathrm{LC}}$ has arbitrary joins, you do not actually even need to explicitly show that it has arbitrary meets.