$f$ is continuous and of bounded variation on $[a,b]$, then $V[a,x]$ is continuous? Can someone help to check my proof? Notation.

Since $f$ is continuous on $[a,b]$, it is uniformly continuous on $[a,b]$. For any $x_0,x_1$ in $[a,b]$, $|f(x_0)-f(x_1)|\to 0$ as $|x_0-x_1|\to 0$.
W.L.O.G., suppose $a\le z_0<z_1\le b$, and let $|z_0-z_1|\to 0$. Consider the partition $P_0$ of $[a,z_0]$ and the partition $P_1$ of $[a,z_1]$. Hence, $\bar{P}=P_0\cup P_1$ is a refinement of $P_1$. Then, we can split the partition $\bar{P}$ by $$A=\bar{P}\cap[a,z_0] \text{ and } B=\bar{P}\cap[z_0,z_1].$$
So, \begin{align*}S_\bar{P}[a,z_1]-S_{P_0}[a,z_0] &=S_A[a,z_0]+S_B[z_0,z_1]-S_{P_0}[a,z_0].
\end{align*}
Note that if $|\bar{P}|\to 0$, then $|P_0|\to 0$. Since $f$ is continuous, $$\lim_{|\bar{P}|\to 0}S_{\bar{P}}[a,z_1] = V[a,z_1] \text{ and } \lim_{|\bar{P}|\to 0} S_{P_0}[a,z_0] = V[a,z_0].$$
Hence, \begin{align*}V[a,z_1]-V[a,z_0] &= V[a,z_0]+V[z_0,z_1]-V[a.z_0]\\
 &= V[z_0,z_1].
\end{align*}
Since $B$ is a partition of $[z_0,z_1]$ with finitely many points and $f$ is uniformly continuous on $[z_0,z_1]$, we conclude that $V[z_0,z_1]\to 0$ as $|z_0-z_1|\to 0$.

I could use $\epsilon$-$\delta$ argument, but I would need to find the precise value of $\delta$ such that $|x_0-x_1|<\delta \implies |f(x_0)-f(x_1)|<\epsilon$. Is this proof valid? Do I need to make any improvement?
 A: The vast majority of your argument up to the equation following "Hence" involves unnecessary and inaccurate reasoning to prove simply that 
$$\tag{*} V[a,z_1] = V[a,z_0] + V[z_0,z_1]$$
For example, the assertion that $|\bar{P}| \to 0$ implies $|P_0| \to 0$ is false. We can have $|P_1| \to 0$ and $|\bar{P}| \to 0$ while $P_0$ stays fixed.  Furthermore, continuity of $f$ is not required to prove (*).
If we take a partition $P$ of $[a,z_0]$ and a partition $P'$ of $[z_0,z_1]$, then $P'' = P \cup P'$ is a partition of $[a,z_1]$ and $S_P[a,z_0] + S_{P'}[z_0,z_1] = S_{P''}[a,z_1]$. 
This implies that $S_P[a,z_0] + S_{P'}[z_0,z_1] \leqslant V[a,z_1]$, and  taking suprema on the LHS we get
$$V[a,z_0] + V[z_0,z_1] \leqslant  V[a,z_1] $$
Another straightforward argument gives us the reverse inequality, thereby proving (*).
With that out of the way, we can move on to the final part of your  argument where you conclude that $V[z_0,z_1] \to 0 $ as $|z_0 - z_1| \to 0$ because $f$ is uniformly continuous and a partition has a finite number of points -- both obviously true but inessential facts.  This is the crux of the proof that the total variation is continuous, and you have simply made an assertion without supplying any details.
For a correct proof, first prove right-continuity by showing that 
$$\lim_{z_1 \to z_0+} V[z_0,z_1] = 0$$ 
By continuity of $f$, given $\epsilon > 0$ there is $\xi > z_0$ such that $|f(z_1) - f(z_0)| < \epsilon/2$ if $z_0 < z_1 < \xi$. Also since total variation is a supremum over partitions, there is a partition $z_0 = x_0 < x_1 < \ldots < x_n = b$ of $[z_0,b]$ such that
$$V[z_0,b] < \sum_{k=1}^n|f(x_k) - f(x_{k-1})| + \epsilon/2$$
If $x_0 = z_0 < z_1 < \min(x_1,\xi)$ then $|f(x_1) - f(x_0)| \leqslant |f(z_1) - f(x_0)| +|f(x_1) - f(z_1)| $
and
$$V[z_0,b] < |f(z_1) - f(z_0)| + |f(x_1) - f(z_1)| + \sum_{k=2}^n|f(x_k) - f(x_{k-1})| + \epsilon/2 \\< \epsilon/2 + |f(x_1) - f(z_1)| + \sum_{k=2}^n|f(x_k) - f(x_{k-1})| + \epsilon/2\\ < V[z_1,b] + \epsilon $$
This implies for all $z_0 < z_1 < \min(x_1,\xi)$ we have 
$$V[z_0,z_1] = V[z_0,b] - V[z_1,b] < \epsilon,$$
which completes the proof of right-continuity.  By a similar argument we can prove left-continuity. 
