Assuming dependent choice, every locally compact Hausdorff space is a Baire space

By [L1] it suffices to show that every nonempty open $W\subseteq X$ meets ${\bigcap }_{n\in \mathbb{N}}{U}_n$; if $X$ has no nonempty open subset the requirement is vacuous and there is nothing to prove, so fix a nonempty open $W$.

$W\cap U_0$ is open and nonempty, $U_0$ being dense and $W$ nonempty open; fixing a point of it, [L2] gives an open $V_0$ with that point in $V_0$ and $\overline{V_0}\subseteq W\cap U_0$, with $\overline{V_0}$ a compact subset of $X$. In particular $V_0$ is nonempty.

Let $\mathcal{A}$ be the set of nonempty open $V\subseteq X$ whose closure $\bar{V}$ is a compact subset of $X$; it contains $V_0$ by step 2.1, so it is nonempty. For $n\in \mathbb{N}$ define a relation $R_n$ on $\mathcal{A}$ by declaring $V<msub><mi>R</mi><mi>n</mi></msub>V^{\mathrm{\prime }}$ to hold exactly when $\overline{V^{\mathrm{\prime }}}\subseteq V\cap U_{n+1}$.

Each $R_n$ is entire on $\mathcal{A}$: given $V\in \mathcal{A}$, the set $V\cap U_{n+1}$ is open and nonempty, $U_{n+1}$ being dense and $V$ nonempty open, so fixing a point of it and applying [L2] gives an open $V^{\mathrm{\prime }}$ containing that point with $\overline{V^{\mathrm{\prime }}}\subseteq V\cap U_{n+1}$ and $\overline{V^{\mathrm{\prime }}}$ compact; then $V^{\mathrm{\prime }}\in \mathcal{A}$ and $V<msub><mi>R</mi><mi>n</mi></msub>V^{\mathrm{\prime }}$. This is a pure existence statement and selects nothing, which is why dependent choice and nothing stronger is spent below.

By [L6] applied to $\mathcal{A}$, the relations $R_n$ and the point $V_0$, there is a sequence $(V_n{)}_{n\in \mathbb{N}}$ in $\mathcal{A}$ with $V_0$ as given and $\overline{V_{n+1}}\subseteq V_n\cap U_{n+1}$ for every $n\in \mathbb{N}$.

The sets $\overline{V_n}$ are nonempty, since $V_n$ is nonempty and $V_n\subseteq \overline{V_n}$ by [L3], and they decrease: $\overline{V_{n+1}}\subseteq V_n\subseteq \overline{V_n}$ by step 5.1 and [L3]. Each is a compact subset of $X$ and hence closed in $X$ by [L3], so each is the trace of a closed set on $\overline{V_0}$ and therefore closed in the subspace $\overline{V_0}$ by [L4].

The family $\{\overline{V_n}:n\in \mathbb{N}\}$ of closed subsets of the compact space $\overline{V_0}$ has the finite intersection property: the intersection of the empty list is $\overline{V_0}$, which is nonempty, and the intersection of a nonempty finite list is $\overline{V_N}$ for $N$ the greatest of the indices occurring, the sets being decreasing, and that is nonempty. So [L5] gives a point $x\in {\bigcap }_{n\in \mathbb{N}}\overline{V_n}$.

That point lies in $W\cap {\bigcap }_{n\in \mathbb{N}}{U}_n$: from $x\in \overline{V_0}\subseteq W\cap U_0$ it lies in $W$ and in $U_0$, and for every $n\in \mathbb{N}$ it lies in $\overline{V_{n+1}}\subseteq V_n\cap U_{n+1}\subseteq U_{n+1}$, so it lies in $U_m$ for $m=0$ and for every $m$ of the form $n+1$, that is in every $U_m$.

So every nonempty open $W$ meets ${\bigcap }_{n\in \mathbb{N}}{U}_n$, which by [L1] makes that intersection dense; as $(U_n)$ was an arbitrary sequence of dense open sets, $X$ is a Baire space.