We are given a continuous semimartingale $Y$ and a continuous process $B$ of finite variation. Hence, we know that $\langle B \rangle$, the quadratic covariation of $B$, is zero in probability.
I now think that $$ \int_0^t Y_s \text{ d}\langle B \rangle_s = 0 \text{ (a.s.)} $$ but I have no clue how I can prove this, or even know that it is indeed true.
Can anyone help me with this?
As you pointed out, we know that $\langle B \rangle_t = 0$ in probability for $t \geq 0$. In fact, $\langle B \rangle_t = 0$ almost surely (mind that the exceptional set may depend on $t$). Indeed: By definition,
$$\langle B \rangle_t = \mathbb{P}-\lim_{|\Pi| \to 0} V_t(B,\Pi)$$
where
$$V_t(B,\Pi) := \sum_{t_k \in \Pi} (B_{t_k}-B_{t_{k-1}})^2.$$
Here $\Pi = \{0=t_0 < \ldots < t_n=t\}$ denotes a partition of the interval $[0,t]$ with mesh size $|\Pi|$. Since
$$|V_t(B,\Pi)| \leq \left( \sup_{|r-s| \leq |\Pi|; r,s \in [0,t]} |B_r-B_s| \right) \sum_{t_k \in \Pi} |B_{t_k}-B_{t_{k-1}}|$$
it follows from the fact that $(B_t)_{t \geq 0}$ has continuous sample paths of bounded variation that $V_t(B,\Pi) \to 0$ almost surely as the mesh size $|\Pi|$ tends to zero. Hence, $\langle B \rangle_t=0$ almost surely.
Now, since $(B_t)_{t \geq 0}$ is continuous, we also know that $(\langle B \rangle_t)_{t \geq 0}$ has continuous sample paths. Hence, by continuity, we get
$$\mathbb{P}(\forall s \in [0,t]: \langle B \rangle_s = 0) = 1$$
for all $t \geq 0$. Now it follows from the very definition of the Riemann-Stieltjes integral that
$$\int_0^t Y_s \, d \langle B \rangle_s = 0$$
almost surely.