In the proof of Krein-Milman in A course in functional analysis by John B. Conway p.142. The following argument is used:
$K$ is a compact convex set in a locally convex space $X$ and $U\subset K$ is a proper convex subset which is open with respect to the subspace topology inherited from $K$. Then in the proof it is claimed that if $x\in U$ and $y\in \overline{U}\setminus U$ then the open line segment from $x$ to $y$ is contained in $U$: that is $(1-t)x+ty\in U$ for $0< t <1$.
However this is not true for any situation. For instance if $K = [-1,1]\times [-1,1]$ and $U = (-1,1)\times (-1,1)\cup (-\frac{1}{2},\frac{1}{2})\times \{1\}$ then $y =(1,1)\in \overline{U}\setminus U$ and $x=(0,1)\in U$ however $(1-t)x+ty\notin U$ for $1/2\leq t<1$.
Can the property that $U$ is a maximal proper open subset of $K$ be used to get around this fact?
I think you're right, this is a gap in Conway's proof. The problem seems to be that he applies Proposition IV.1.11, which would require $x$ to be an interior point of $U$ (with respect to the ambient topology of $X$); this would be valid if $U$ were open in $X$, but here it is only relatively open in $K$. This error doesn't appear in the errata list on his web page, so you might like to let him know about it.
To fix his proof, let's follow his notation and let $T_{x,\lambda}(y) = \lambda y + (1-\lambda) x$ for $x,y \in K$ and $0 \le \lambda < 1$. For $x \in U$, the set $T^{-1}_{x,\lambda}(U)$ is a relatively open convex subset of $K$ that contains $U$, so by maximality it is either $K$ or $U$.
We want to show it is $K$. Suppose not; then there exists $y \in K$ with $T_{x,\lambda}(y) \notin U$. Then we can find some other point $z$ on the line joining $x,y$ such that $z \notin U$ but $T_{x,\lambda}(z) \in U$. For example, you could set $z_0 = y$, $z_{n+1} = T_{x,\lambda}(z_n)$. Then $z_n \to x$, and so since $U$ is relatively open, we have $z_n \in U$ for sufficiently large $n$. Now let $z = z_k$ with the greatest $k$ such that $z_k \notin U$. This would show $T_{x,\lambda}^{-1}(U) \ne U$, contrary to our assumption.
Intuitively, the idea is that if the open line from $x$ to $y$ were not contained in $U$, we could dilate the set $U$ outward from $x$, just enough to make it contain some additional points of the line but still exclude $y$. That would produce a larger convex relatively open proper subset of $K$.