...it is not true that we have a distinguished element with norm zero. @Pietro. I don't know what you mean by saying that "A linear normed space is naturally a metric space." Well, I know that, for $(V, \|\cdot\|)$ a normed space over a given valuated field $(\mathbb K, |\cdot|)$, the map $d: |V| \times |V| \to \mathbb{R}_0^+: (x,y) \mapsto \|x-y\|$ is a metric on $|V|$ (the carrier of $V$). What I don't really understand is why $d$ is pretended to be natural (which is, in fact, part of another question that I'd like to raise). That said, one of many possible follow ups... (tbc)

...contains, among the other marvelous insights, the key observation that symmetry is not an essential property of metrics. Today in hindsight, it may sound trivial. But, given that Lawvere's research grew up on the path carved out by a somewhat consolidated tradition, let me say that it was indeed a small Copernican revolution. Yes, I could actually reply that, but I won't. Instead, I will invite you to consider that normed semigroups, in spite of the fact that, for some apparent reasons, very few people have studied them, are an example of algebraic normed structures for which... (tbc)

@Yemon. Nah, wait! First, "Common things with norms" is not the same as "Things with norms": this is cheating! :) Second, you know much better than me that, in the search of a unified view pointing to glue together ideas so far from each other (at least apparently), one has to separate what is "essential" from what is rather "accidental". Based on this, I could actually reply your question with another question: Why on earth should "things with norms" be inherently characterized by the property of having one distinguished element with norm zero? Lawvere's work on generalized metrics... (tbc)

No, I don't agree. It would be more correct to say that some things with norms have a distinguished point, namely the element of norm zero.

I see, nothing but the (strict) pullback of $F_1$ and $F_2$. Ok!

Nothing more specific than this? Ok, I will wait to hear from somebody else, if any, before accepting your answer. Thanks.

@Yulia. Yes, I need it to be linearly ordered.

@Yulia. Too bad that your nice counterexample doesn't work also for the 2nd question. In any case, I'm going to edit the OP and add that you replied Q1 in the negative.

On another hand, Gillman and Jerison's 1960 result addressed by problem 2.1.J on page 74 states a kind of extensibility characterization applying, in a cumulative sense, to every continuous function from a topological subspace of $\mathcal{X}$ to the interval [0,1] (endowed with the obvious topology). P.S.: to be sure we're talking about the same things, let me mark that I'm referring herein to the 1989 Heldermann-Verlag edition of Engelking's General Topology.

@Yulia. Thanks for the reference. Clearly, up to the trivial case in which I is finite (possibly after removing all duplicates), $\{X_i\}_{i \in I}$ won't generally satisfy the condition of being a locally finite cover. All the more that Engelking's Prop. 2.1.13 (pp. 72-73) applies to locally finite closed covers, as you noticed in your last comment. Nevertheless, $\{X_i\}_{i \in I}$ isn't a completely arbitrary cover of $\mathcal{X}$, here: it's a chain in $(2^X,\subseteq)$ and each of its members are dense in $\mathcal{X}$. So I don't see how this could imply a negative answer to Q1.

@fedja. Obvious?! I'd say it's a great remark. @Michael. You're clearly right. I'm going to edit the OP according to the comments of you both and withdraw what is actually unnecessary.