Q:

Path Connectedness and fixed points

We have the following given to us,
Let $α, β \colon [0, 1] \to [0, 1]$ be (not necessarily continuous) functions such that

$α(x) ≤ β(x)$, for all $x ∈ [0, 1]$.
The set $K = \{\,(x, y); α(x) ≤ y ≤ β(x)\,\}$ is closed in $\mathbb R^2$
.

Show that there exists $t ∈ [0, 1]$ such that $(t, t) ∈ K$.
We think that if $K$ is closed, then it must be path-connected. In which case, we have a continuous function from $[0,1]$ to $[0,1]$ that is contained in $K$. Therefore can use the intermediate value theorem to show that this function must have a fixed point, which would conclude our argument.
Our difficulty is that we are unable to prove that $K$ is actually path-connected, even though every example of $α$ and $β$ that we can think of satisfying the required properties make $K$ path connected. 

A:

Let $A=\{\,x\in[0,1]:\alpha(x)>x\,\}$. Then $A$ is an open subset of $[0,1]$: If $\alpha(x_0)>x_0$, then $K$ is disjoint to an open neighbourhood of the compact set $\{x_0\}\times[0,x_0]$, hence $\alpha(x)>x$ in a neighbourhood of $x_0$. 
Similarly, $B=\{\,x\in[0,1]:\beta(x)<x\,\}$ is open. Since $A\cap B=\emptyset$ and $[0,1]$ is connected, we conclude that $A\cup B\ne [0,1]$. Let $t\in[0,1]\setminus (A\cup B)$. Then $\alpha(t)\le t\le\beta(t)$, i.e. $(t,t)\in K$.

