A function lattice with the two-point duplication property uniformly approximates its target

Fix $x\in X$ and let $H_x:=\{h\in L:h(x)=f(x)\}$; for $h\in H_x$ put $U_h:=\{z\in X:h(z)>f(z)-\epsilon \}$, an open set by continuity.

The family $(U_h{)}_{h\in H_x}$ covers $X$: for any $y\in X$, [L1] supplies $h\in L$ with $h(x)=f(x)$ and $h(y)=f(y)>f(y)-\epsilon$, so $h\in H_x$ and $y\in U_h$.

By compactness and [L2], finitely many $U_{h_0},\dots ,U_{h_n}$ cover $X$; their pointwise maximum $g:=h_0\vee \cdots \vee h_n$ belongs to $L$, satisfies $g(z)>f(z)-\epsilon$ for every $z\in X$, and satisfies $g(x)=f(x)$ because every $h_j(x)=f(x)$.

Let $G:=\{g\in L:g(z)>f(z)-\epsilon \text{ for every }z\in X\text{, and }g(x)=f(x)\text{ for some }x\in X\}$, a subset of $L$ formed by comprehension rather than by selecting one function per point, and for $g\in G$ put $V_g:=\{z\in X:g(z)<f(z)+\epsilon \}$; each $V_g$ is open. The family $(V_g{)}_{g\in G}$ covers $X$: given $x\in X$, step 3.1 produces a member of $L$ with both defining properties, so it lies in $G$, and it contains $x$ in its $V_g$ because $g(x)=f(x)<f(x)+\epsilon$.

By compactness and [L2], finitely many $V_{g_0},\dots ,V_{g_m}$ cover $X$; their pointwise minimum $u:=g_0\wedge \cdots \wedge g_m$ belongs to $L$.

Every $g_j$ is greater than $f-\epsilon$ everywhere by step 3.1, so $u>f-\epsilon$ everywhere; and at every $z$ some $V_{g_j}$ contains $z$, so $u(z)\le g_j(z)<f(z)+\epsilon$. Thus $\mathrm{\mid }u(z)-f(z)\mathrm{\mid }<\epsilon$ for every $z\in X$.