You are confusing the measure on path space with Lebesgue measure. The "almost everywhere" refers to the former: almost every individual path can be taken to be continuous everywhere. Indeed, the Wikipedia page you link to says that Brownian motion is "almost surely everywhere continuous".
In other words, if $\mathbb{P}$ is Wiener measure on a suitable measurable space $(\Omega, \mathcal{F})$, then there is a set $N \subset \Omega$ of $\mathbb{P}-$measure $0$ such that for all $\omega \notin N$, $t \rightarrow \omega(t)$ is continuous for all $t \in [0,1]$.
We have to assume that the underlying probability space is complete; otherwise the assertion might fail.
So, suppose that $(\Omega,\mathcal{A},\mathbb{P})$ is a complete probability space and $(X_t)_{t \in [0,1]}$ a process with almost surely continuous sample paths, i.e. there exists a null set $N \in \mathcal{A}$ such that $$[0,1] \ni t \mapsto X_t(\omega)$$ is continuous for all $\omega \in \tilde{\Omega} := \Omega \backslash N$. Now
$$\tilde{X}_t(\omega) := \begin{cases} X_t(\omega), & \omega \in \tilde{\Omega}, \\ 0, & \omega \in N \end{cases}$$
defines a stochastic process on $\Omega$ with continuous sample paths, and therefore
$$\sup_{t \in [0,1]} \tilde{X}_t = \sup_{t \in [0,1] \cap \mathbb{Q}} \tilde{X}_t$$
is measurable as countable supremum of measurable random variables. On the other hand, we have
$$\tilde{S}_t(\omega) = \sup_{t \in [0,1]} \tilde{X}_t(\omega) = \sup_{t \in [0,1]} X_t(\omega)= S_t(\omega) \quad \text{for all $\omega \in \tilde{\Omega} = \Omega \backslash N$}$$
and so
$$\{S_t \in B\} = \left( \{\tilde{S}_t \in B \} \cap N^c \right) \cup \bigg( \{S_t \in B \} \cap N \bigg)$$
for any Borel set $B$. Since $N \in \mathcal{A}$ and $\tilde{S}_t$ is measurable, we know that
$$\left( \{\tilde{S}_t \in B \} \cap N^c \right) \in \mathcal{A}.$$
Moreover,
$$\left\{ S_t \in B \right\} \cap N \subseteq N$$
and since the probability space is complete, this implies
$$\left\{ S_t \in B \right\} \cap N \in \mathcal{A}.$$
Combining both considerations proves $\{S_t \in B\} \in \mathcal{A}$, and this proves the measurability of $S_t$.
Remark More generally, the following statement holds true in complete probability spaces:
Let $(\Omega,\mathcal{A},\mathbb{P})$ and $(E,\mathcal{B},\mathbb{Q})$ be two measure spaces and assume that $(\Omega,\mathcal{A},\mathbb{P})$ is complete. Let $X, Y: \Omega \to E$ be two mappings. If $X$ is measurable and $X=Y$ almost surely, then $Y$ is measurable.
Best Answer
Since there are uncountably many $t$'s, it is not automatic that something that is almost surely true for any given $t$ is almost surely true for all $t$. For example, consider a Poisson process $N(t)$. For each $t$, $N(t)$ almost surely doesn't have a jump at $t$. On the other hand, with probability $1$ the process does have a jump somewhere.