This is my current understanding of the weak topology. Let $V$ be a vector space and $V^*$ its dual. Then the weak topology on $V$ is the coarsest topology such that the evaluation maps $$v \mapsto f(v), \quad \forall f \in V^*.$$ are continuous. Further, a sequence $\{v_i\} \in V$ is said to converge weakly to $v \in V$ if $$f(v_i) \rightarrow f(v) \quad \forall f \in V^*.$$
I have two questions about the above. The first, according to Lax's book on functional analysis the open sets of the weak topology are unions of finite intersections of sets of the form $$\{x : a < f(x) < b\},$$ but this is not immediately obvious to me. Is this because we take some neighborhood of $f(x)$, which intuitively contain all values between $a$ and $b$ and we simply take the preimage? This may be pedantic, but the field over which $V$ exists, call it $\mathbb{K}$, was never assumed to be ordered so how can we conclude this?
Second, in order to determine weak convergence we look at sequences of the form $f(v_i) \rightarrow f(v)$, in other words we are determining convergence over the $\mathbb{K}$. However, similar to before, $\mathbb{K}$ was assumed to simply be a field, not necessarily a complete field, so there is no metric. How is this convergence calculated?
Most of the functional analysis is implicitly (or not) assumed to be over $\mathbb{R}$. Some of it is assumed to be over $\mathbb{C}$, but not in this particular case, where we use ordering.
Note that convergence does not require a metric. And indeed, some people do consider general topological fields, but this is a rather esoteric maths. The main field of study is always over reals $\mathbb{R}$ or complex numbers $\mathbb{C}$.