Let $K$ be compact, convex and metrizable in the topological vector space $X$ (local convexity isn't required for the argument). Fix a metric $d$ on $K$ compatible with the topology. Let $F_n = \left\{x \in K\,:\, \text{there are } y,z \in K\text{ such that }x = \frac{1}{2}(y+z)\text{ and }d(y,z) \geq \frac{1}{n}\right\}.$ Then $F_n$ is closed and a point is non-extremal if and only if it is in the $F_\sigma$-set $F = \bigcup_n F_n$. Thus the set of extremal points $\partial_e K = K \smallsetminus F$ is a $G_\delta$.
Note: I posted the above as a comment in this related MO thread where it is also mentioned that this may fail if $K$ is compact convex but not metrizable (even if $X$ is assumed to be locally convex): the set $\partial_e K$ need not even be a Borel set in $K$ by an example described by Bishop–de Leeuw in section VII of The representations of linear functionals by measures on sets of extreme points. Annales de l'institut Fourier, 9 (1959), p. 305–331.
Added:
Some more details:
The set $C_n = \{(y,z) \in K \times K\,:\,d(y,z) \geq 1/n\}$ is closed since it is the preimage of $[1/n,\infty)$ under the continuous function $d\colon K \times K \to [0,\infty)$. Therefore $C_n$ is compact. The set $F_n$ is the image of the compact set $C_n$ under the continuous function $(y,z) \mapsto \frac{1}{2}(y+z)$, hence $F_n$ is compact and thus closed.
We have $\partial_e K = K \smallsetminus F$ where $F = \bigcup_n F_n$ as above.
If $x$ is not an extremal point, then $x \in F_n$ for some $n$:
By definition $x = (1-\lambda) y + \lambda z$ for some $0 \lt \lambda \lt 1$ and $y \neq x \neq z$. If $\lambda = \frac{1}{2}$ then $x \in F_n$ where $n$ is so large that $\frac{1}{n} \leq d(y,z)$.
If $\lambda \neq \frac{1}{2}$ we may (possibly after switching $y$ and $z$) assume that $0 \lt \lambda \lt \frac{1}{2}$ and write $x = \frac{1}{2}(y + z_\lambda)$ where $z_\lambda = (1-2\lambda)y + 2\lambda z \in K$. Since $y \neq z$ we have $y \neq z_\lambda$ and thus $d(y,z_\lambda) \geq \frac{1}{n}$ for large enough $n$, so $x \in F_n$.
Conversely, if $x \in F_n$ for some $n$ then $x$ is clearly not extremal.
Later:
Let me add some remarks on your approaches:
The problem with idea (1) is that $U_{\lambda}$ is not open. In fact, I showed in point 2. above that $U_{1/2} = \partial_e K$ and a small modification of that argument gives that $U_{\lambda} = \partial_e K$ for all $\lambda \in (0,1)$, so you're right that $\partial_e K = \bigcap_{\lambda} U_{\lambda}$, but proving that $U_{\lambda}$ is a $G_\delta$ is the same as the original problem, so that idea won't lead anywhere.
The second idea looks much better, however I doubt that exploiting separability of $K$ only is enough (that is: I doubt that the set of extremal points of a compact convex separable but non-metrizable set is a $G_\delta$, but I haven't checked this thoroughly). I think the argument I gave is one way to get around the difficulties.
Another point I'd like to add is that the distinction of boundary points and interior points you seem to be making does not work for infinite-dimensional compact convex sets. In fact, the Hilbert cube $C$ is homogeneous in the sense that its homeomorphism group acts transitively (see here for a good write-up of the non-trivial proof), so no point is distinguished by topological properties alone. This property is generic in the sense that every compact convex metrizable set in a locally convex vector space is either contained in a finite-dimensional subspace or homeomorphic to $C$ by a theorem of Klee. See my answer here for more on this.
Finally, we can't hope to do much better than $G_\delta$. In three dimensions take the double cone $K$ obtained by taking the convex hull of a circle $C$ of radius $1$ around $(1,0,0)$ in the $(x,y)$-plane and the two points $p_\pm=(0,0,\pm1)$ on the $z$-axis. Then $\partial_e K = \{p_\pm\} \cup C\smallsetminus \{(0,0,0)\}$ is not closed. [In two dimensions the non-extremal points are open in the boundary, hence the set of extremal points of a compact convex set is closed, the one-dimensional case is trivial.]