Continuity of Kakutani fixed points

Let $$X$$ be a compact and convex space and let $$T=[0,1]$$ be some parameter space. Let $$F:X\times T\rightrightarrows X$$ be a correspondence that is compact-valued, convex, and upper-hemicontinous. By Kakutani's fixed point theorem, there is a fixed point $$x(t)=F(x(t),t)$$ for each parameter $$t\in T$$.

Suppose we also know that

1. the set $$F(x, t')$$ converges $$F(x, t)$$ as $$t'\to t$$ for any $$x\in X$$ (in the sense that the indicator functions for the two sets converge pointwise), and
2. there is a unique fixed point $$x(t)$$ for each $$t\in T$$.

Question: Is this enough to determine that $$x(t)$$ is a continuous function?

Here is my very informal attempt at a proof: Consider sequences $$(x_n, t_n)\to (x, t)$$ such that $$x_n\in F(x_n, t_n)$$. Suppose $$x\notin F(x, t)$$. Since, $$F(\cdot, t)$$ and $$F(\cdot, t_n)$$ are close for large enough $$n$$, then $$x\notin F(x,t_n)$$. By UHC of $$F$$, for any open set $$V$$ containing $$F(x, t_n)$$, I can find an open set $$U$$ of $$x$$ such that for all $$x'\in U$$, $$F(x', t_n)\in V$$. I can pick $$V$$ such that it excludes $$x$$, and by consequence, any $$x_n$$ for $$n$$ large enough. However, this would imply that there is an $$x_n\in U$$ and $$x_n\notin V$$, which implies $$x_n\notin F(x_n, t_n)$$. Hence, $$x\in F(x, t)$$, and the graph $$\{(x,t): x\in F(x, t) \}$$ is closed. This along with the uniqueness should be sufficient for showing that $$x(t)$$ is a continuous function.

I assume that you mean that $$F$$ is upper semicontinuous on the product space. Then in particular (since $$X$$ is compact, Hausdorff and $$F$$ has closed values), $$F$$ has a closed graph. This implies that also the set $$\{(x,t):x\in F(x,t)\}$$ is closed. This means that the multimap $$t\mapsto\{x:x\in F(x,t)\}$$ has a closed graph and assumes closed values. Since $$X$$ is a compact space, it follows that this multimap is upper semicontinuous.
In particular, your hypothesis 1 superfluous. (Well, actually it follows from the upper semicontinuity of $$F$$ with respect to both variables - it is the special case that you fix one variable.) Note that you really have to require the upper semicontinuity with respect to both variables, even in the metric case: Your first part of the proof becomes wrong unless you assume some sort of locally uniform convergence (which implies of course again the upper semicontinuity).
• I meant for $F$ to be upper hemicontinuous only in $x$ but I see that my proof does require uniform convergence. Thanks!