Is it true that $ \limsup_{T\to0+}F(0,t)=\lim_{T\to0+}\left[\sup_{0\le s<t\le T}F(s,t)\right]$?

I have a proof, but in the comments there is a counterexample: where is the problem?

The problem is here

we get $$ \lim_{n\to+\infty}F(s_n^{(\delta_n)},t_n^{(\delta_n)}) =\lim_{T\to0+}\left[\sup_{0\le s<t\le T}F(s,t)\right]=c\; $$

The convergence for each fixed $n$ of a sequence $a_{m,n}$ to a limit $c_n$ when $m$ tends to infinity does not necessarily implies that a diagonal sequence $a_{n,n}$ converges to a $\lim c_n$ when $n$ tends to infinity. For instance, we can put $a_{m,n}=0$ when $m=n$ and $a_{m,n}=1$, otherwise.

The statement "Then by continuity, $\forall \epsilon>0$, $\exists N_\epsilon$..." is wrong. If a function $F$ is continuous on an open set and $\xi_n$ is a sequence converging to a boundary point then $F(\xi_n)$ need not be Cauchy (which in fact is the claim you make). For the argument to hold you need uniform continuity, but then $F$ extends continuously to $(0,0)$ which is not what you are interested in. A concrete example is: $ F(s,t) = \frac{s}{t}$ for $0\leq s<t$ and $F(s,t) = \frac{t}{s}$ for $0\leq t < s$. You have $\limsup_T F(0,T)=0$ and $\limsup_{s,t} F(s,t)=1$.

The application you have in mind is, however, somewhat different. So let $$ F(s,t) = \frac{f(s)-f(t)}{|s-t|^\lambda}$$ with $f$ an $\alpha$-Hölder continuous function.

If $ 0 < \lambda < \alpha \leq 1 $ then $F$ is in fact uniformly continuous, since you have: $$ |F(s,t) | \leq \frac{ C|t-s|^\alpha}{|t-s|^\lambda} = C |t-s|^{\alpha-\lambda}.$$ so there is no problem.

In the case $0<\lambda=\alpha\leq 1$ the conclusion is, however, wrong. To constuct a counter-example, fix $k>0$ and define for $A>0$ the spike of amplitude $A$: $$ \phi_A(x) = \max\{0, A - k |x|^\alpha\}$$ This non-negative function is bounded by $A$, has support in $|x|\leq (A/k)^{1/\lambda}$ and has $\alpha$-Hölder constant $k$ (independent of $A$): $$ \sup_{x\neq y} \frac{|\phi_A(x)-\phi_A(y)|}{|x-y|^\alpha} = k .$$

Now, define $$ f(t) = \sum_{n\geq 1} \phi_{\frac{1}{n^{2\lambda}}} (t-\frac{1}{n}) .$$ The spikes have disjoint support for $k$ large enough and $f(1/n)=1/n^{2\lambda}$ so the function verifies: $f(t)\leq t^{2\lambda}$, $t\geq 0$. In particular, $$ \limsup_{T\rightarrow 0^+} F(0,T) = \limsup_T f(T)/T^\lambda \leq C \limsup_T T^{2\lambda-\lambda} = 0. $$ On the other hand we have when $\alpha=\lambda$: $$ \limsup_{T\rightarrow 0^+} \sup_{0<t<s\leq T} F(t,s)= \limsup_T \sup_{0<t<s\leq T} \frac{|f(s)-f(t))|}{|s-t|^\lambda} = k $$ since arbitrarily close to the origin we have a spike with Hölder constant $k$. By letting $k$ increase with $n$ you may also construct an example where the first limsup is zero and the last infinity. In other words in order to obtain the result you need better control on the function $f$ when approaching the origin.

According to one of your comments elsewhere on this page it looks as if you may be interested in the case $\alpha=\lambda=1/2$ and $f(t)$ being e.g. a standard Brownian motion. In that case I believe that a.s. both limits are infinity (so in a sense you are then right). It is well-know that the $\limsup_{T}\sup_{s,t}$ is infinity. This follows for example from Lévy's modulus of continuity. see e.g. Peres and Morters, Thm 1.14. I believe that is also true a.s. for the point wise limit $\limsup_T F(0,T)$ but couldn't find a direct proof. There exists points where the limit is finite (sometimes called $\alpha$ slow times) but it seems that they have zero Lebesgue measure, although possibly Hausdorff dim 1 (?). But I am far from sure about this last point.