Fewest steps to reach $200$ from $1$ using only $+1$ and $×2$

Look at what the operations $[+1]$ and $[\times 2]$ do to the binary expansion of a number:

• $[\times 2]$ appends a $0$, and increases the length by one, leaving the total number of $1$'s unchanged;
• if the final digit is $0$, then $[+1]$ increases the number of $1$'s by one, but doesn't change the length;
• if the final digit is $1$, then $[+1]$ doesn't increase the total number of $1$'s (it may in fact decrease it), and doesn't increase the total length by more than $1$.

Therefore, with a single key press:

• you can increase the length by one, but this won't increase the number of $1$'s;
• you can increase the number of $1$'s by one, but this won't increase the length.

The binary expansion of $200$ is $200_{10}=11001000_2$. This has three $1$'s, and a length of eight. Starting from $1$, we must increase the length by seven, and increase the number of $1$'s by two. So this requires at least nine steps.

You can proceed by induction on $n$, showing that the quickest way to reach any even number $2n$ involves doubling on the last step, which is clearly true for the base case $n=1$ (where doubling and adding $1$ have a tomato-tomahto relationship).

Now if the last step to reach $2n+2$ isn't doubling, it can only be adding $1$ from $2n+1$. But $2n+1$ can only be reached by adding $1$ from $2n$, at which point the inductive hypothesis says the next previous number was $n$. But you can get from $n$ to $2n+2$ more quickly in two steps: add $1$, then double. So the last step in the quickest route to $2n+2$ is doubling from $n+1$.

Setting the display to binary base, $[\times2]$ inserts a $0$ to the right and $[+1]$ increments; if the rightmost digit is a zero, it just turns it to a $1$.

Using these rules you build a number of $o$ ones and $z$ zeroes in $o-1+z$ keystrokes, starting from $1$. This seems close to optimal.