If $f(x_i)<c$ for all $i$, then $$w=\lim_{i\to \infty } f(x_i)\leq c.$$
Indeed, if $w>c$, then there is $i$ s.t. $c<f(x_i)<w$ which is a contradiction with $f(x_i)\leq c$ for all $i$.
In general, if a sequence is s.t. $x_n< \ell$ for all $n$, then $$\lim_{n\to \infty }x_n\leq \ell.$$
In your proof, since $f(x_i)\leq c$ for all $i$, if $w>c$, then $$f(x_i) <c\implies -c\leq -f(x_i)\implies 0 <w-c<w-f(x_i) \leq |f(x_i)-w|.$$
Therefore, $$\lim_{i\to \infty }|f(x_i)-w|\geq w-c>0,$$
which contradict that $f(x_i)\underset{n\to \infty }{\longrightarrow} w$.
Indeed, if $w>c$, then there is $i$ s.t. $c<f(x_i)<w$ which is a contradiction with $f(x_i)\leq c$ for all $i$.
In general, if a sequence is s.t. $x_n< \ell$ for all $n$, then $$\lim_{n\to \infty }x_n\leq \ell.$$
In your proof, since $f(x_i)\leq c$ for all $i$, if $w>c$, then $$f(x_i) <c\implies -c\leq -f(x_i)\implies 0 <w-c<w-f(x_i) \leq |f(x_i)-w|.$$ Therefore, $$\lim_{i\to \infty }|f(x_i)-w|\geq w-c>0,$$ which contradict that $f(x_i)\underset{n\to \infty }{\longrightarrow} w$.