Let $b>a\ge0$, $$\mathfrak T_{[a,\:b]}:=\left\{\left(t_0,\ldots,t_k\right):k\in\mathbb N\text{ and }a=t_0<\cdots<t_k=b\right\}$$ and $$\left|\varsigma\right|:=\max_{1\le i\le k}\left(t_i-t_{i-1}\right)\;\;\;\text{for }\varsigma=\left(t_0,\ldots,t_k\right)\in\mathfrak T_{[a,\:b]}\;.$$ Moreover, let $f:[a,b]\to\mathbb R$, $$V_\varsigma(f):=\sum_{i=1}^k\left|f(t_i)-f(t_{i-1})\right|\;\;\;\text{for }\varsigma=\left(t_0,\ldots,t_k\right)\in\mathfrak T_{[a,\:b]}$$ and $$V_{[a,\:b]}(f):=\sup_{\varsigma\in\mathfrak T_{[a,\:b]}}V_\varsigma(f)\;.$$
Now, assume $V_{[a,\:b]}(f)<\infty$ and $f$ is continuous. Let $(\varsigma_n)_{n\in\mathbb N}\subseteq\mathfrak T_{[a,\:b]}$ with $$\left|\varsigma_n\right|\xrightarrow{n\to\infty}0\;.\tag1$$ I want to show that $$V_{\varsigma_n}(f)\xrightarrow{n\to\infty}V_{[a,\:b]}(f)\tag2\;.$$
By definition, $$V_{\varsigma_n}(f)\le V_{[a,\:b]}(f)\;\;\;\text{for all }n\in\mathbb N\;.\tag3$$ Assume that $(2)$ doesn't hold. Then there is an $\varepsilon>0$ with $$\forall N\in\mathbb N:\exists n\ge N:V_{\varsigma_n}(f)\le V_{[a,\:b]}(f)-\varepsilon\tag4\;.$$ Moreover, by definition of the supremum, there is a $\varsigma\in\mathfrak T_{[a,\:b]}$ with $$V_{[a,\:b]}(f)-\varepsilon<V_\varsigma(f)\tag5\;.$$
Now, I don't know how I need to proceed. Clearly $|f(a)-f(b)|\le|f(a)-f(t)|+|f(t)-f(b)|$ for all $t\in[a,b]$ and hence $V_\varsigma(f)\le V_{\varsigma\cup\varsigma_n}(f)$, but I don't know how I could use this to obtain a contradiction. So, how can we proceed?
Let $\varsigma = (x_0, x_1, \dotsc, x_m) \in \mathfrak{T}_{[a,b]}$ and $\varepsilon > 0$.
By the uniform continuity of $f$, choose $0 < \eta < \frac{1}{2} \min \{ x_k - x_{k-1} : 1 \leqslant k \leqslant m\}$ such that $\lvert f(x) - f(y)\rvert \leqslant \frac{\varepsilon}{2m}$ whenever $\lvert x-y\rvert \leqslant \eta$.
For $\pi = (t_0,t_1,\dotsc,t_r) \in \mathfrak{T}_{[a,b]}$ with $\lvert \pi\rvert < \eta$, let $\ell_{\mu} = \max \{ t_{\rho} : t_{\rho} \leqslant x_{\mu}\}$ and $u_{\mu} = \min \{ t_{\rho} : t_{\rho} \geqslant x_{\mu}\}$ for $0 \leqslant \mu \leqslant m$. Then
$$V_{\pi}(f) \geqslant \sum_{\mu = 1}^{m} \lvert f(\ell_{\mu}) - f(u_{\mu-1})\rvert.$$
Since $\lvert \ell_{\mu} - x_{\mu}\rvert < \eta$ and $\lvert u_{\mu} - x_{\mu}\rvert < \eta$, we have $\lvert f(x_{\mu}) - f(\ell_{\mu})\rvert \leqslant \frac{\varepsilon}{2m}$ and $\lvert f(x_{\mu}) - f(u_{\mu})\rvert \leqslant \frac{\varepsilon}{2m}$. Hence
$$\lvert f(\ell_{\mu}) - f(u_{\mu-1})\rvert \geqslant \lvert f(x_{\mu}) - f(x_{\mu-1})\rvert - 2\frac{\varepsilon}{2m}$$
for all $\mu$, and thus
$$V_{\pi}(f) \geqslant V_{\varsigma}(f) - \varepsilon$$
for all $\pi \in \mathfrak{T}_{[a,b]}$ with $\lvert \pi\rvert < \eta$. Therefore we have
$$\liminf_{n\to\infty} V_{\varsigma_n}(f) \geqslant V_{\varsigma}(f) - \varepsilon.$$
Since $\varepsilon > 0$ and $\varsigma \in \mathfrak{T}_{[a,b]}$ were arbitrary, it follows that
$$\liminf_{n\to\infty} V_{\varsigma_n}(f) \geqslant V_{[a,b]}(f).\tag{1}$$
Since
$$\limsup_{n\to\infty} V_{\varsigma_n}(f) \leqslant V_{[a,b]}(f)\tag{2}$$
is clear, we have
$$\lim_{n\to\infty} V_{\varsigma_n}(f) = V_{[a,b]}(f)\tag{3}$$
as desired.