Why is $x$ free but $\forall x$ is not?

$F(x)$ is open because it only gets a truth value once you decide what $x$ is.

$\forall x.F(x)$ is closed because it already has a truth value. You don't need to decide on any particular value for $x$ to ask whether $F(x)$ is true for all values.

For example $x+1=3$ is sometimes true (such as when $x$ is $2$) and sometimes false (such as when $x$ is $27$).

But $\forall x. x+1=3$ is simply false. You can't meaningfully ask whether $\forall x.x+1=3$ is true or not when $x$ is $27$, because the truth value of $\forall x.x+1=3$ does not depend on any external mechanism to supply a value for $x$. In particular you cannot make $\forall x.x+1=3$ become true by declaring that $x$ is $2$, because the formula itself specifies what it means by $x$.

True or false?

$$\forall x. x+1 > x$$

"True, of course. There is no counter example."

True or false?

$$\forall x. x\ge 0$$

"False, of course. $x$ could be $-17$, which is a counter example."

True or false?

$$x\ge 0$$

"How should I know? You didn't say what $x$ is."