If $f:[0,\infty) \to \mathbb{R}$ is a continuous function and $D \subset [0,\infty)$ a dense set, then $$\forall s \in [0,t]: f(s) \geq -a \iff \forall s \in [0,t] \cap D: f(s) \geq -a.$$ The implication "$\Rightarrow$" is obvious and the other implication follows from the continuity of $f$ (proof by contradiction: if we had $f(s)<-a$ for some $s \in [0,t]$, then by the continuity of $f$ we could find $s' \in [0,t] \cap D$ such that $f(s')<-a$.)
Applying this equivalence for the sample paths of Brownian motion, we find that $$\{\forall s \in [0,t]: B_s \geq -a\} = \bigcap_{s \in [0,t] \cap D} \{B_s \geq -a\}.$$ If we consider the sequence of refining partitions $\Pi_j = \{t_1^{(j)}<\ldots<t_{n_j}^{(j)}\}$ from your question, then this gives
$$\{\forall s \in [0,t]: B_s \geq -a\} = \bigcap_{s \in [0,t] \cap D} \{B_s \geq -a\}= \bigcap_{j \geq 1} A_j$$
where $$A_j := \bigcap_{k=1}^{n_j} \{B_{t_k} \geq -a\}.$$ Since the sequence of partitions is refining, the sets $A_j$ are decreasing in $j$. Hence, by the continuity of the measure
$$\mathbb{P}(\forall s \in [0,t]: B_s \geq -a) = \lim_{j \to \infty} \mathbb{P}(A_j) = \lim_{j \to \infty} \mathbb{P} \left( \bigcap_{k=1}^{n_j} \{B_{t_k} \geq -a\} \right).$$ Since the fnite-dimensional distributions are (non-degenerate) Gaussian, they do not have point masses, and so
$$\mathbb{P}(\forall s \in [0,t]: B_s \geq -a)= \lim_{j \to \infty} \mathbb{P} \left( \bigcap_{k=1}^{n_j} \{B_{t_k} > -a\} \right). \tag{1}$$
Finally, note that
$$\mathbb{P}(\forall s \in [0,t]: B_s \geq -a) = \mathbb{P}(\forall s \in [0,t]: B_s > -a) \tag{2}$$
which follows, for instance, from the reflection principle or from the strong Markov property of Brownian motion (see this answer). Combining $(1)$ and $(2)$ proves the desired identity.
