Proof that if $\phi \in \mathbb{R}^X$ is continuous, then $\{ x \mid \phi(x) \geq \alpha \}$ is closed. Recently, having realized I did not properly internalize it (shame on me!), I went back to the definition of continuity in metric spaces and I found a proposition for which I was looking for a proof.
Here there is the result and my "proof" (in the hope to get rid of the quotation marks).
[In general, I use $N_{\varepsilon, X} (x)$ to denote an open $\varepsilon$-neighbourhood of $x \in X$.]

Proposition: Let $\phi \in \mathbb{R}^X$ be a continuous function, with $X$ an arbitrary metric space. Then, the set $\{ x \mid \phi(x) \geq \alpha \}$ is closed.
Attempted proof:
Let $\alpha$ be an arbitrary real number. We establish the result by showing that $\{ x \mid \phi(x) < \alpha \}$ is an open set in $X$. Notice that for every $x \in X$, if $\phi (x) < \alpha$, then there is a $\varepsilon > 0$ such that there is an open neighbourhood $N_{\varepsilon, \mathbb{R}} (\phi (x)) < \alpha$. Let $z \in X$ be arbitrary and such that $\phi (z) < \alpha$. Hence, by the  definition of continuity and the fact that $\phi$ is continuous, there is a $\delta (z, \varepsilon) > 0$ such that
$$N_{\delta, X} (z) \subseteq \phi^{-1} ( N_{\varepsilon, \mathbb{R}} ( \phi (z)). $$
Hence, being $z \in X$ arbitrary, the proposition follows.

Is this proof correct?
As always, any feedback is more than welcome.
Thank you for your time.

Edit:
I know it is possible to proceed, as hinted by air in a comment below, through the fact that the if a function is continuous, then the preimage of a closed set is closed. However, I find this solution a bit too topological, in the sense that I really would like to know about this $\varepsilon - \delta$ proof, which – to me – has a stronger metric flavour.
 A: I'm not sure why you need an $\epsilon$-$\delta$ proof when we aren't trying to prove continuity.  We are just trying to prove $\{x \mid \phi(x) \geq a \}$ is closed if $\phi$ is continuous.
It suffices to show the complement is open.  But since $\{x \mid \phi(x) \geq a \} = \phi^{-1}( [a, \infty) )$, and $f^{-1}(B^{c}) = (f^{-1}(B))^{c}$ for every function $f$ and every set $B$ in the codomain of $f$, we have:
$[a,\infty)^{c} = (-\infty, a)$ which is open.  Since $\phi$ is continuous, $\phi^{-1}( (-\infty, a))$ is open.  But $\phi^{-1}( (-\infty, a)) = \phi^{-1}( [a, \infty)^{c}) = (\phi^{-1}([a, \infty))^{c}$.
Thus, $(\phi^{-1}([a, \infty))^{c}$ is open, and so its complement, $\phi^{-1}([a,\infty))$ is closed. But $\phi^{-1}([a,\infty))$ is just another notation for $\{x \mid \phi(x) \geq a \}$, so our set is closed, as desired.
A: Ok now the proof is basically correct (when we are working in metric spaces)! Some remarks:
As Umberto P. also noted in his answer in the related question you asked, I am not fond of the notation "$N_{\varepsilon, \mathbb{R}} (\phi (x)) < \alpha$". In fact in the other thread you write "$\phi(Y) \le \alpha$" for a set $Y$, which is still a bit more appropriate than what you write here (though still false).
You should write instead: There is a neighborhood $N_{\varepsilon, \mathbb{R}} (\phi (x))$ such that or all $y \in N_{\varepsilon, \mathbb{R}} (\phi (x))$ it holds that $y < \alpha$.
Also towards the end, your argument is correct but I usually like to be a bit more explicit (at least until you get more comfortable with the material). For example I would write: By the continuity of $\phi$ there exists $\delta:=\delta (z, \varepsilon) > 0$ such that $|\phi(x)-\phi(z)| < \varepsilon$ for all $x \in N_{\delta, X}(z)$. This implies that for $x \in N_{\delta, X}(z)$ it holds that $\phi(x) \in N_{\varepsilon, \mathbb{R}} (\phi (z))$. Therefore we finally have that:
$$N_{\delta, X} (z) \subseteq \phi^{-1} ( N_{\varepsilon, \mathbb{R}} ( \phi (z)) \subseteq \{ x \mid \phi(x) < \alpha \}$$
This finishes our proof. (Note the final inclusion; you had already shown it but I still feel it is critical to the argument and should be repeated at this point.)
