Question Regarding Proposition 8.1.18 in Engelking Proposition 8.1.18 in Engelking is stated without proof, and I am having trouble proving the second part of the assertion. It would be greatly appreciated if I received a hint. Note that Engelking assumes his uniformities to be Hausdorff, which is not completely standard, so I've inserted the word Hausdorff where it's not stated in Engelking's book.  First, I state two properties that are needed to state the proposition: 
Suppose $X$ is a set and $P$ is a collection of pseudometrics on $X$. Say $P$ satisfies (UP1) and (UP2) if the following are satisfied: 
$\text{(UP1)}$ If $\rho_1, \rho_2 \in P$, then $\max\{\rho_1, \rho_2\} \in P;$ 
$\text{(UP2)}$ For every pair $x,y$ of distinct points of $X$, there exists a $\rho \in P$ such that $\rho(x,y)>0$. 
Note that if a family of pseudometrics satisfies (UP2), the uniformity which is generated by this family is Hausdorff; (UP1) implies that the collection $\{V_{d, \varepsilon} \}_{\varepsilon>0, d}$ is a base and not merely a subbase for some uniformity on a set $X$, where $V_{d, \varepsilon} : = \{(x,y) | d(x,y)<\varepsilon\}$. 
One more piece of terminology: given a uniformity $\mathcal U$ on a set $X$ and a pseudometric $\rho$ on $X$, say $\rho$ is uniform with respect to $\mathcal U$ if for every $\varepsilon>0$, there is $U \in \mathcal U$ such that $\rho(x,y) < \varepsilon$ whenever $(x,y) \in U$. This condition is equivalent to the uniform continuity of $\rho$ on the product uniform space $X \times X$. 
I now state the proposition, slightly paraphrased. 
Proposition 8.1.18 (Engelking) Suppose we are given a set $X$ and a family $P$ of pseudometrics on $X$ which has properties $\text{(UP1) - (UP2)}$. The family $\mathcal B$ of all entourages of the diagonal which are of the form $V_{\rho, 2^{-i}}$ for $\rho \in P$ and $i \in \mathbb N$ is a base for a (Hausdorff) uniformity $\mathcal U$ on the set $X$. Moreover, every pseudometric $\rho \in P$ is uniform with respect to $\mathcal U$. 
If, moreover, $(X, \tau)$ is a topological space and all pseudometrics of the family $P$ are continuous as maps $X \times X \to [0, \infty)$, and if for every $x \in X$ and every nonempty closed set $A \subseteq X$ such that $x \notin A$ there exists a $\rho \in P$ such that $\inf\{\rho(x,a) | a \in A\} >0$, then $\mathcal U$ is a (Hausdorff) uniformity on the space $X$. 
Note: following Engelking's terminology, if $X$ is a topological space and $\mathcal U$ is a uniformity on $X$, and if the uniform topology induced by $\mathcal U$ coincides with the given topology on $X$, say that $\mathcal U$ is a uniformity on the space $X$. Thus, the second paragraph in the proposition above is saying that the uniform topology induced by $\mathcal U$ (let's call it $\tau_{\mathcal U}$) coincides with $\tau$. 
Could I have a hint to prove the assertion in the second paragraph of the proposition? (I see how to prove the assertions in the first paragraph). With notation as before, I've only been able to show $\tau \supseteq \tau_{\mathcal U}$; this follows readily from the continuity of the pseudometrics on the product space $X \times X$. 
In fact, I'm having trouble believing why this is true. Consider the following example: 
Let $X= \{f \in C^1(\mathbb R) | ~f(0)=0\}$, let $d_{n, k}(f,g):= \sup_{|x| \leq n} |D^k f(x)-D^kg(x)|$, $k=0,1$. Let $\tau$ be the topology generated by $\{d_{n, k}\}_{n \in \mathbb N, k=0,1}$. Let $P:= \{d_{n, 1}\}_{n}$, and let $\mathcal U$ and $\tau_{\mathcal U}$ be the uniformity generated by $P$ and the uniform topology, respectively. Then, 


*

*Every $d_{n, 1}$ is continuous as a map $X \times X \to [0, \infty)$; 

*$P$ satisfies properties (UP1) and (UP2); to see (UP2), suppose $f,g \in X$ satisfy $d_{n,1}(f,g) = 0$ for every $n$; that is, $f'=g'$. Since $f(0)=g(0)$, it follows that $f=g$. 

*The second assumption in the second paragraph of Prop. 8.1.18 is satisfied: suppose $A \subseteq X$ is $\tau$ closed and suppose $f \notin A$. For the sake of contradiction suppose that for every $n$, there is a sequence $(g_{n,k})_k$ in $A$ such that $d_{n, 1}(f, g_{n,k}) \to 0$ as $k \to \infty$. That is, $g_{n,k}' \to f'$ uniformly on $[-n, n]$, and $(g_{n,k}(0))_k$ of course converges; hence by a well known theorem of real analysis, $(g_{n,k} |_{[-n, n]})_k$ converges uniformly to some differentiable $g_n: [-n,n] \to \mathbb C$, $g_n(0)= 0$, and $g_n' = f'|_{[-n, n]}$ (and hence $g_n = f|_{[-n, n]}$). Thus, given $\varepsilon>0, n \in \mathbb N, k \in \{0,1\}$, there is $h \in A$ such that $d_{n, k}(f, h) < \varepsilon$. Consequently every $\tau$ neighborhood of $f$ meets $A$, hence $f\in A$. 


But, $\tau$ is strictly finer than $\tau_{\mathcal U}$. 
I am having doubts about my "counterexample", so it would also be appreciated if someone could point out any errors I made in my example.
 A: I found the mistake in my “counterexample”: it is not true that $\tau$ is strictly finer than $\tau_{\mathcal U}$, because if $f_j’ \to g$ for some $g \in X$ locally uniformly, then it follows that there is $f$ such that $f_n \to f$ locally uniformly and $f$ is continuously differentiable (by the fact that the $f_j $ are continuously differentiable) and $f’=g$. (This all holds because the sequence $(f_j(0))_j$ converges - indeed, it’s just the constant sequence). In other words, $d_{n,1}(f_j, g) \to 0$ for each $n$ implies $d_{n, 0}(f_j, g) \to 0$ for each $n$. 
Next, I have a proof now of the reverse inclusion that $\tau \subseteq \tau_{\mathcal U}$. Fix $x \in X$, and take an (open) $\tau$ neighborhood $N$ of $x$. Then $N^c$ is closed, and for some $\rho \in P$ and some $\varepsilon>0$, we have $\rho(x, a)>\varepsilon$ for every $a \in N^c$ (this follows by our assumption with $A=N^c$). Hence $N^c \subseteq (V_{\rho, \varepsilon}[x])^c$, so $x \in V_{\rho, \varepsilon}[x] \subseteq N$, and so the $\tau_{\mathcal U}$ neighborhood filter of $x$ refines the $\tau$ neighborhood filter of $x$.  
Finally, if we do not assume $\text{(UP2)}$ in the theorem above, all the conclusions remain true if we omit the word Hausdorff (by the same proof).
