In my analysis class we were given the following problem:
Let $(f_n)_{n\in \mathbb{N}}$ be a sequence of continuous functions in $[0,1]$ and assume they converge uniformly to some $f:[0,1] \to \mathbb{R}$. If $(x_n)_{n\in \mathbb{N}}$ is a sequence in $[0,1]$ and $x_n \rightarrow x$ show that $f_n(x_n) \xrightarrow{} f(x)$.
The tutor solved it using the triangle inequality in this form
$$ |f_n(x_n)-f(x)| \leq |f_n(x_n)-f(x_n)| + |f(x_n)-f(x)| \leq ||f_n-f||_{\infty}+ |f(x_n)-f(x)|$$ and to make this small he used uniform convergence of $(f_n)_{n\in \mathbb{N}}$ and the continuity of $f$ along with convergence of $(x_n)_{n\in \mathbb{N}}$.
This is all fine. Afterwards he said it would be a good exercise to try and understand why we can't a priori use the triangle inequality in this way $$ |f_n(x_n)-f(x)| \leq |f_n(x_n)-f_n(x)| + |f_n(x)-f(x)| $$ to solve the exercise, because we wouldn't be able to make $|f_n(x_n)-f_n(x)|$ arbitrarily small, without going back to the other way.
Is this really true?
The problem is that to show $f_n(x_n)\to f_n(x)$, we have to use the fact $f_n\to f$ uniformly, which leads us back to the first triangle inequality.
To elaborate a little more:
As $n$ increases, both $x_n$ and $f_n$ are changing, thus we cannot guarantee $f_n(x_n)\to f_n(x)$ unless there is some uniform convergence of $f_n$, here is an example:
Let $f_n(x)=x^n$, $f(x)=0$ if $x<1$ and $f(x)=1$ if $x=1$, then $f_n\to f$ not uniformly on $[0,1]$. Consider $x=1$ and $x_n=1-1/n$, then $x_n\to x$ but
$$f_n(x_n)-f_n(x)=(1-1/n)^n-1\to1/e-1\neq0\ \ \text{as}\ n\to\infty.$$
This shows we cannot prove $f_n(x_n)\to f_n(x)$ without knowing $f_n\to f$ uniformly.
If we do know $f_n\to f$ uniformly, then
$$|f_n(x_n)-f_n(x)|\leq|f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|+|f(x)-f_n(x)|\to0,$$
but this is basically the idea of the first triangle inequality in you post.