The answer is no to both questions. E.g., suppose that $T=2$ and 
$$f_n(t)=1(t\le1-\tfrac1n)+n(1-t)1(1-\tfrac1n<t\le1).$$
Then $f_n(t)$ increases to $f(t)=1(t<0)$, and all the other conditions on $f_n$ hold. 

However, by the [definition of the Skorokhod metric][1], for $t_n:=1-\tfrac1{2n}$ and any strictly increasing continuous function $h\colon[0,2]\to[0,2]$ we have  
$$d(f_n,f)\ge|f_n(t_n)-f(h(t_n))|=|\tfrac12-f(h(t_n))|
\ge\tfrac12,$$
since $f(t)\in\{0,1\}$ for each $t\in[0,2]$. So, $d(f_n,f)\not\to0$. 


  [1]: https://en.wikipedia.org/wiki/C%C3%A0dl%C3%A0g#Skorokhod_space