In the $x + 1$ problem, does every positive integer $x$ eventually reach $1$? I know that the more famous $3x + 1$ problem is still unresolved. But it seems to me like the similar $x + 1$ problem, with the function
$$f(x) = \begin{cases} x/2 & \text{if } x \equiv 0 \pmod{2} \\ x + 1 & \text{if } x \equiv 1 \pmod{2}.\end{cases}$$
should be very easy to prove it with just my modest, haphazardly obtained knowledge of mathematics. But the only thing I can think of is "of course every $x$ has to reach $1$," which is clearly not a rigorous proof.
Is this easy to prove, or is it perhaps just as difficult as the $3x + 1$ problem?

The variant you have described is not hard. For example one can give a formal proof by strong induction.
Here is the induction step. Suppose that we ultimately end up at $1$ if we start at any number $u\lt k$. We show that we end up at $1$ if we start at $k$.
If $k$ is even, in one step we are below $k$, and we are finished.
If $k$ is odd and greater than $1$, in two steps we are at $\frac{k+1}{2}\lt k$.
