Is this proof correct? I did it myself so I'm not sure if it's right.
Let ${f_k}$ be a sequence of bounded variation on $[a,b]$ such that for each $x\in [a,b]$ $f(x)=\lim_{k\to \infty}f_k(x)$. If there exist $K>0$ such that $V(f_n,[a,b])\le K$ for all $n\in \mathbb N$, then $f$ is bounded variation on $[a,b]$.
Proof:
Let $\epsilon >0$.
Then $V(f,[a,b])=\sum_{i=1}^{k}|f(x_i)-f(x_{i-1})|=\sum_{i=1}^{k}|\lim_{n\to \infty}f_n(x_i)-\lim_{n\to \infty}f_n(x_{i-1})|=\sum_{i=1}^{k}\lim_{n\to \infty}|f(x_i)-f(x_{i-1})|=\lim_{n\to \infty}\sum_{i=1}^{k}|f(x_i)-f(x_{i-1})|=\lim_{n\to \infty}K=K$.
Therefore $f$ is bounded variation on $[a,b]$.
Please tell me if it is correct or if it's not.
The idea is correct but there are a few mistakes. The variation is the supremum over all $\sum_i^k |f(x_i)-f(x_{i-1})|$ as you change the partitions so you cannot write $V(f,[a,b])=\sum_i |f(x_i)-f(x_{i-1})|$. Start with a given partition $a=x_0<\cdots<x_k=b$. Then $$\sum_i^k |f(x_i)-f(x_{i-1})|=\lim_{n\to\infty}\sum_i^k |f_n(x_i)-f_n(x_{i-1})|.$$ Since the variation is the supremum you have $$\sum_i^k |f_n(x_i)-f_n(x_{i-1})|\le V(f_n,[a,b])\le K$$ for each $n$ and so $$\lim_{n\to\infty}\sum_i^k |f_n(x_i)-f_n(x_{i-1})|\le K.$$ This shows that $$\sum_i^k |f(x_i)-f(x_{i-1})|\le K$$ for every partition $a=x_0<\cdots<x_k=b$ of $[a,b]$. Hence, $K$ is an upper bound and so by the definition of supremum of a set you get $V(f,[a,b])\le K$.