Problem. Suppose that $F$ is an uncountable algebraically closed field and $D$ is a division $F$-algebra. If $\dim_F D$ is countable, then $D=F$.
For each $x\in D$, $F(x)$ is a field extension of $F$ and therefore $[F(x)\colon F]\leqslant\aleph_0$. Since $F$ is algebraically closed, if $F(x)$ is a proper extension, it must be transcendental. Thus I think if the uncountableness of $F$ implies that the degree of any transcendental simple extension must be uncountable, then the problem is done, and this is where I am stuck. So I would like to ask whether I am in a right way, and what to do for the rest of the proof? Thanks in advance..
Elaborating on the linear independence of the set described in the comments above, namely for some fixed $x\notin F$, $X = \left\{\frac{1}{x-a}\mid a\in F\right\}$. Note that since $x$ is trancendental over $F$, we may as well assume that $F(x)$ is the field of rational functions with coefficients in $F$ and variable $x$.
Take some linear combination of the elements of $X$ that equals $0$. It is of the form $$ 0 = \frac{b_1}{x-a_1} + \frac{b_2}{x-a_2}+\cdots\frac{b_n}{x-a_n} $$ where all the $a_i$ are distinct. Multiply this by $\prod_{i = 1}^n(x-a_i)$, and we get $$\begin{align} 0 = {}&b_1(x-a_2)(x-a_3)\cdots(x-a_n) \\ &{}+ b_2(x-a_1)(x-a_3)\cdots(x-a_n)\\ &\vdots\\ &{}+b_n(x-a_1)(x-a_2)\cdots(x-a_{n-1}) \end{align}$$ Since this is an element in the subring of polynomials of the field of rational functions, we have evaluation homomorphisms $h_a: F[x]\to F$, defined by $h_a(f) = f(a)$. Now note what happens when we apply $h_{a_i}$ to the element above. It becomes $0 = b_i(a_i-a_1)\cdots(a_i-a_n)$ (where the term $(a_i-a_i)$ is skipped). Since all the $a_i$ were distinct, this means that $b_i = 0$, and thus the linear combination above is trivial (all coefficients $b_i$ are $0$), so $X$ is linearly dependent.