I've done a bit of digging but I haven't found any sources to confirm that this is true.
Take $\limsup{\{x_n\}}=L$ and $f(x): D \to \mathbb{R}$ to be a continuous and monotonically increasing function such that $\{x_n\} \in D$. Does $\limsup{f(x_n)} = f(L)$?
I take it to be true because $f$ preserves order, and as there are only a finite number of terms greater than a given $L + \epsilon$ then after the application of the function there will still only be a finite number of terms greater than $f(L) + \epsilon$.
Any help would be appreciated!
Quoting your question:
"Take $\limsup{\{x_n\}}=L$ and $f(x): D \to \mathbb{R}$ to be a continuous and monotonically increasing function such that $\{x_n\} \in D$. Does $\limsup{f(x_n)} = f(L)$?"
i.) Given that $f(x)$ is continuous and monotonically increasing has no logical connection to $\{x_n\} \in D$.
ii.) I am assuming that your question refers to $D$ bounded.
Thus:
Using Heine-Borel theorem: $lim_{x\rightarrow L}f(x)$ exists and $lim_{x \rightarrow L}f(x)=M$ iff for every sequence $x_n\rightarrow L, x_n\neq L, \forall n\implies f(x_n)\rightarrow M$.
Now:
$f(x)$ is continuous in $D\implies lim_{x\rightarrow L}f(x)=f(L)=M$ by definition of continuity.
which is equivalent to: $\forall (x_n)\rightarrow L \implies f(x_n)\rightarrow M$
Thus: Let $(x_n)$ be a sequence with $limsup(x_n)=L$.
Then there exists a subsequence of $x_n$ Namely, $x_{n_k}$, s.t. $x_{n_k}\rightarrow L$.
And since $f(x)$ is continuous $lim f(x_{n_k})\rightarrow M$.