General fact: if $X$ is a set and we have two topologies $\mathcal{T}$ and $\mathcal{T}'$ on it, and $\mathcal{T}$ has a base $\mathcal{B}$ and $\mathcal{T}'$ has a base $\mathcal{B}'$, then $\mathcal{B}=\mathcal{B}'$ implies $\mathcal{T}=\mathcal{T}'$.
This is what Munkres means by that having the same base means having the same topology. The proof is immediate from definitions. But if you're sadistically minded, you can check lemma 13.3 for two inclusions, but you don't need it (a set is open iff it's a union of base sets etc.)...
He applies this to $\mathcal{T}=\mathcal{T}_d$ and $\mathcal{T}'=\mathcal{T}_{\bar{d}}$, taking
$\mathcal{B} = \{B_d(x,r) \mid x \in X,0 <r< 1\}$ and
$\mathcal{B}' =\{B_{\bar{d}}(x,r) \mid x \in X, 0 <r< 1 \}$
Using the general fact he mentions in the first line, applied to both $d$ and $\bar{d}$:
If $(X,\rho)$ is any metric space, $\mathcal{B} = \{B_\rho(x,r) \mid x \in X,0 <r< 1\}$ is a base for $\mathcal{T}_\rho$.
Proof: if $O$ is open in $\mathcal{T}_\rho$ and $x \in O$, there is some $r'>0$ so that $B_\rho(x,r') \subseteq O$. But define $r = \min(\frac12, r') < 1$ and note that $B=B_\rho(x,r) \in \mathcal{B}$ and $x \in B \subseteq B_\rho(x,r') \subseteq O$ so that we've verified $\mathcal{B}$ is a base for $\mathcal{T}_\rho$ and we're done.
As the defined collections $\mathcal{B}$ and $\mathcal{B}'$ are identical, the first fact finishes the proof.
QED. This is Munkres' argument in unnecessary detail.
First I will give an alternative proof:
Note that by Theorem 26.6, you don't actually need to directly check that $f^{-1}$ is continuous, it is enough to show that $f$ is a continuous bijection.
The function $f$ is continuous because it is an isometry. Indeed, for any $\epsilon > 0$, choose $\delta = \epsilon$. Then if $d(x,y) < \delta$, we have $d(f(x),f(y)) = d(x,y) < \delta = \epsilon$.
An isometry is immediately injective. As for subjectivity, assume to the contrary that there is an $a\in X$ such that $a\not\in f(X)$. Then there is an $\epsilon$-neighborhood around $a$ disjoint from $f(X)$. Define the sequence $(x_n)$ as in the hint. Since $x_k\in f(X)$ for all $k>1$ we have $d(x_1,x_k) \geq \epsilon$. Fix some positive integer $n$ and let $k = m - n + 1$ and apply the fact that $f$ is an isometry $n-1$ times. The above then becomes $d(x_n, x_m) \geq \epsilon$ whenever $n\neq m$.
Finally, since $X$ is a compact metric space, it is sequentially compact. Thus the sequence $(x)_n$ has a convergent subsequence. However, as shown above the distance between any two points in this subsequence must be larger than or equal to $\epsilon$, thus it cannot converge. This contradiction shows $f$ must be surjective.
All in all, $f$ is a continuous bijection, and per the first paragraph, $f$ is a homemorphism.
Now I will comment on your proof(s):
A general comment is that you overuse symbols, which makes the text difficult to read. This looks more like a proof sketch which was copied, rather than a completed proof. Even if completely logically correct, proofs written in this style are difficult to parse. Write in complete sentences. I looked at some of your other questions and this is a recurring situation.
A second general comment is that you do not always need to specify logical quantifiers if the context is obvious. In particular, your repeated use of the "for all" symbol.
All in all, your proof needs a lot of work. It's a mess to read. I'll try and cover it in detail, but keep in mind these comments are also comments on your general style, and do not just apply to these specific instances. I will also not suggest alternative strategies in this proof, since I covered that above. Thus I will only focus on clarifying the proof using your proof strategy.
let f(x)=f(x′), where x,x′∈X.
You do not need to specify where $x$ and $x'$ come from, it is obvious from context.
Then d(f(x),f(x′))=d(x,x′)=0 by isometry property.
This is better expressed as "since $f$ is an isometry". It's also not a correct English sentence. It would at least have to be "the isometry property".
Now we show f is continuous. Let x∈X. Given ϵ>0. Take δ=ϵ. If ∀y∈X with d(x,y)<δ , then d(f(x),f(y))=d(x,y)<ϵ. Thus f is continuous, ∀x∈X.
This is a mess. You simply have to show that if two distinct points are delta-close, their images are epsilon-close. You instead fix your $x$, and check every $y$, which is equivalent, but a lot more messy. The penultimate sentence is also malformed, since it reads: "If for all y in X with the distance from x to y being less than delta, then ...". Again, there is that overuse of the "for all" symbol at the end.
Now we show f is surjective. Assume towards contradiction, f(X)≠X. So ∃a∈X such that a∈X−f(X).
The context is already clear here. You do not need to specify that $a\in X - f(X)$. Simply saying that $a \not\in f(X)$ is enough.
By theorem 26.5, f(X) is compact. By theorem 26.3, f(X) is closed in X. So X−f(X)∈TX.
It's generally better to write "is open" rather than use set membership.
∃ϵ>0 such that Bd(a,ϵ)⊆X−f(X).
Do not start a sentence with mathematical symbols. Also, instead of using symbols here you should simply write "there is an epsilon-ball around $a$ disjoint from $f(X)$. It's the exact same statement as what you wrote, but is much easier to parse.
It’s easy to check xn+1∈f(X), ∀n∈N by induction.
First, you do not need to specify where the $n$s live. It is clear from context. Second, this is not shown by induction. Every point $x_k$ lives in $X$, so its image will live in $f(X)$. You can use induction, but it is overkill.
Since Bd(a,ϵ)∩f(X)=∅, xn+1∉Bd(a,ϵ),∀n∈N.
Please, oh please stop cramming "∀n∈N" in everywhere. The reader understood it the first time, and would have understood it the zeroth time. This is the last time I will comment on it.
Claim: ∀k∈N, d(xk,xn)≥ϵ ∀n∈N∖{k}.
Generally inserting claims in the middle of proofs is bad style. Not always, but generally. It is bad style in this case. What you wrote is also much clearly expressed as "$d(x_k, x_n) \geq \epsilon$ whenever $k\neq n$. Again, logically equivalent but easier to read.
Proof: Base case, k=1.
Explicitly labelling base cases and inductive steps is not something that is done.
Inductive step: Assume d(xl,xn)≥ϵ ∀n∈N∖{l}; l∈N. Then d(xl+1,xn)=d(f(xl),f(xn−1))=d(xl,xn−1)≥ϵ,∀n∈N∖{l+1}, last inequality follows from inductive hypothesis.
I know what you're trying to do here, but writing $l\in\mathbb{N}$ makes it seem like you've shown it for all $l$ already, which you haven't. Also, you reference $n-1$ which might be 0, which is an issue. This needs to be fixed.
Which implies xn≠xm, if n≠m.
It does, but you never use this, so why include it?
By theorem 28.2, X is sequentially compact. So {xn} have convergent subsequence.
Spelling error. have -> has
Let {xni} is subsequence of {xn}
This should be "Let ... be a subsequence".
such that {xni}→x0∈X. For ϵ/2, ∃N∈N such that d(xni,x0)<ϵ, ∀i≥N. Then d(xnN,xnN+1)≤d(xnN,x0)+d(x0,xnN+1)<ϵ. So d(xnN,xnN+1)<ϵ; nN<nN+1.
You explicitly choose epsilons here, which is fine, but a bit overkill. You already know different points in the sequence are "far" apart, so its impossible for any subsequence to converge, for the subsequence will consist of point that are "far" apart.
Now we show f−1 is continuous. Since f is bijective, f is invertible. So ∃!g:X→X such that f∘g=idX=g∘f. Given ϵ>0. Take δ=ϵ. Let x,y∈X with d(x,y)<δ. Since x,y∈X, ∃i,j∈X such that f(i)=x and f(j)=y. Then d(g(x),g(y))=d(g(f(i)),g(f(j)))=d(i,j)=d(f(i),g(j))=d(x,y)<ϵ. Thus g=f−1 is uniformly continuous. Which implies g is continuous. Hence f is bijective and homeomorphism.
This is mostly fine, if a bit wordy. It is certainly uniformly continuous, but you only need continuity, so settle for that. You do not need to mention uniform continuity.
Your proof is logically correct, but a mess in every other way. Now on to approach 2.
In proof of f is surjective, we can also contradict definition of limit point compact in place of sequential compact. By theorem 28.1, X is limit point compact. ∀k∈N, d(xk,xn)≥ϵ,∀∈N∖{k}. So ∀n,m∈N with n≠m, we have d(xn,xm)≥ϵ>0. Which implies xn≠xm, if n≠m.
I'm assuming here you're using the same strategy as in the first approach to show this? It's not clear exactly.
So E={xn|n∈N} is an infinite subset of X. By definition of limit point compact, ∃x∈X such that x∈E′. ∀V∈Nx,V∩E≠∅. Let V=Bd(x,ϵ/2)∈Nx. By theorem 17.9, |Bd(x,ϵ/2)∩E|=∞. So ∃p,q∈Bd(x,ϵ/2)∩E such that p≠q. Then d(p,q)≤d(p,x)+d(x,q)<ϵ and p,q∈E. So d(p,q)<ϵ,p≠q. Thus we reach contradiction. Rest of the proof is similar to approach(1).
This seems correct, yes. Again, horrible to parse. You need to define your symbols.
As for approach 3, it is correct, although you still need to show surjectivity, injectivity, and continuity of $f$. You only really save on showing that the inverse is continuous.
Best Answer
The two approaches are basically the same:
You assume non-continuity, so some open $O \subseteq Y$ so that $f^{-1}[O]$ is not open, so there is some $x$ with $f(x) \in O$ so that all balls around $x$ "stick out of" $f^{-1}[O]$. In fact what you need here in the proof is that $x$ has a countable local base of neighhbourhoods $U_n$ (we can assume it to be decerasing in $n$, as in the metric case, where we use $U_n = B_d(x,\frac1n)$ as the local base) so that the positive consequence of non-interior-ness of $x$ allows us to pick $x_n \in U_n \setminus f^{-1}[O]$ giving us a sequence converging to $x$ but the $f(x_n) \notin O$ so these do not converge to $f(x)$, which then contradicts the given of sequential continuity, so the assumption of non-contuity was wrong.
Only non-continuity gives us a concrete point to work with to start contradicting sequential continuity. It's just in the nature of the definitions and their logical structure, as it were.
There is some work in the foundations of logic and constructism (where proofs from contradiction are not allowed) to show that continuity and sequential continuity are really logically different. In many branches of constructive maths (like intuitionism) you cannot even define a non-continuous function.
So yes, your proofs are correct, and follow the schedule I described. The converse can be extended even beyoud first countable spaces, to the realm of so-called sequential spaces (defined by the property that a sequentially closed set $F$ is always closed; $F$ is ssequentially closed when for all sequences $x_n$ with terms in $F$ we have $x_n \to x$, we can say $x \in F$, so it contains the limits of all its sequences). Try to generalise the converse to such spaces $X$. All first countable (and thus metric) spaces are sequential, but there are many more.