In general if $X$ is a semimartingale and $H$ is a locally bounded predictable process, then $$
\Delta \left(\int_0^{\cdot} H_s\,\mathrm dX_s\right)_t=H_t \Delta X_t,\quad t\geq 0,
$$
so if $X$ is continuous, then so is any integral with respect to $X$. Now, integration with respect to the Lebesgue measure is just integration with respect to the semimartingale $X_t=t$ (which actually is of finite variation). Since $\Delta X_t=t-t=0$ we have that this $X$ is continuous and hence the integral is as well.
The integral is also of finite variation because the following holds for the stochastic integral:
If $X$ is a semimartingale and $H$ is locally bounded and predictable, then $(\int_0^\cdot H_s\,\mathrm d X_s)_{t\geq 0}$ is also a semimartingale. If $X$ is a local martingale then so is $(\int_0^\cdot H_s\,\mathrm d X_s)_{t\geq 0}$, and lastly, if $X$ is of finite variation, then so is $(\int_0^\cdot H_s\,\mathrm d X_s)_{t\geq 0}$.
This can be seen in Jacod and Shiryaev's Limit Theorems for Stochastic Processes for example.
Actually, the proof is indeed similar to the proof of $(1)$. It's based on the fact that convergence in probability implies almost sure convergence of a subsequence. Set
$$\tau_n := \inf\left\{t \geq 0; \int_0^t |\sigma(s)|^2 \, ds + \int_0^t |b(s)|^2 \, ds \geq n \right\}.\tag{4}$$
By Doob's inequality, Itô's isometry and Tschbysheff inequality we have
$$\begin{align} & \quad \mathbb{P} \left( \sup_{t \leq T} \left| \int_0^t (f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s)) \, dB_s \right| > \varepsilon \right) \\ &\leq \mathbb{P} \left( \sup_{t \leq T} \left| \int_0^{t \wedge \tau_n} (f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s)) \, dB_s \right| > \varepsilon, \tau_n > T \right) + \mathbb{P}(\tau_n \leq T) \\ &\leq \frac{4}{\varepsilon^2} \cdot \underbrace{\mathbb{E} \left( \int_0^{T \wedge \tau_n} |f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s)|^2 \, ds \right)}_{=:I} + \mathbb{P}(\tau_n \leq T) \end{align} $$
(Note that the boundedness of $f'$ implies the existence of the integrals.) Since \begin{align*} f'(X_s^{\Pi}) \cdot \sigma^{\Pi}(s) - f'(X_s) \cdot \sigma(s) &= f'(X_s^{\Pi}) \cdot \big(\sigma^{\Pi}(s)-\sigma(s)\big) \\ &\quad + \sigma(s) \cdot \big(f'(X_s^{\Pi})-f'(X_s) \big)\end{align*} and $(a+b)^2 \leq 2a^2+2b^2$ we obtain $$\begin{align*} I &\leq 2 \mathbb{E} \bigg( \int_0^{T \wedge \tau_n} \underbrace{|f'(X_s^{\Pi})|^2}_{\leq \|f'\|^2_{\infty}} \cdot |\sigma^{\Pi}(s)-\sigma(s)|^2 \, ds \bigg)\\ &\quad + 2 \mathbb{E} \left( \int_0^{T \wedge \tau_n} \sigma^2(s) \cdot |f'(X_s^{\Pi})-f'(X_s)|^2 \, ds \right) \end{align*}$$
The first addend converges to $0$ as $|\Pi| \downarrow 0$ since $\sigma^{\Pi} \cdot 1_{[0,\tau_n)} \to \sigma \cdot 1_{[0,\tau_n)}$ in $L^2(\lambda_T \times \mathbb{P})$ by assumption. For the second one, we note that
\begin{align*} \int_0^{T \wedge \tau_n} \sigma^2(s) |f'(X_s^{\Pi})-f'(X_s))|^2 \, ds &\leq \sup_{s \leq T}|f'(X_s^{\Pi})-f'(X_s))|^2 \int_0^{\tau_n} |\sigma(s)|^2 \, ds \\ &\leq n \sup_{s \leq T}|f'(X_s^{\Pi})-f'(X_s))|^2 \end{align*}
Since $X^{\Pi} \to X$ uniformly in probability, it follows from the continuity of $f'$ that the right-hand side converges to $0$ in probability as $|\Pi| \to 0$. Moreover, by the above estimate,
$$\int_0^{T \wedge \tau_n} \sigma^2(s) |f'(X_s^{\Pi})-f'(X_s))|^2 \,d s \leq 2n \|f'\|_{\infty} \in L^1(\mathbb{P})$$
and so Vitali's convergence theorem gives
$$ \mathbb{E} \left( \int_0^{T \wedge \tau_n} \sigma^2(s) \cdot |f'(X_s^{\Pi})-f'(X_s)|^2 \, ds \right) \to 0$$
Similarily, one can prove the convergence of the other addends in the right-hand side of $(2)$.
Best Answer
Note that the processes $$X(t,\omega)^+ := \max\{0,X(t,\omega)\} \qquad \qquad X(t,\omega)^- := \max\{0,-X(t,\omega)\}$$ are predictable and
$$\int_0^t X(r) \, dr = \int_0^t X_r^+ \, dr - \int_0^t X_r^- \, dr$$
Both integrals on the right-hand side are increasing in $t$, thus $t \mapsto \int_0^t X(r) \, dr$ is of bounded variation.
Suppose that $X$ has càdlàg sample paths. Let $\omega \in \Omega$, $T>0$ and $$c := \sup\left\{\int_0^s \left|X(r,\omega)\right| \, dr ; s \in [0,T]\right\}<\infty.$$
Then, by the mean value theorem,
$$\begin{align*} \left|\exp \left(\int_0^t X_r \, dr \right) - \exp \left( \int_0^s X_r \, dr \right) \right| = \exp(\xi) \cdot \left| \int_0^t X_r \, dr - \int_0^s X_r \, dr \right| \end{align*}$$
for any $s,t \in [0,T]$ and some $\xi=\xi(\omega) \in [-c,c]$. Thus
$$ \left|\exp \left(\int_0^t X_r \, dr \right) - \exp \left( \int_0^s X_r \, dr \right) \right| \leq e^c \cdot \left| \int_0^t X_r \, dr - \int_0^s X_r \, dr \right|$$
Consequently, the claim follows from the fact that $[0,T] \ni t \mapsto \int_0^t X(r) \, dr$ is of bounded variation.