First of all, lets present a strategy for $N=4$.
$N=4$
$O$ can win in 7 moves (13 total).
To start, the first two moves are arbitrary due to rotation and stretching, so let us assume we have the following;
1. $O_1=[0,0]$
2. $X_1=[-1,-1]$
Now, $O$ will pick a line not on the $O_1,X_1$ line and place another point. $X$, meanwhile, should place a point between these two since not doing so give $O$ more options. The placement for $X$ is mostly arbitrary so long as it is between $O$s points.
3. $O_2=[2,0]$
4. $X_2=[1,0]$
At this point, $O$ will start on a different line, 90 degrees from the first;
5. $O_3=[0,2]$

$X$ now has two sets of 2 to worry about; $\{O_1, O_3\}$, and $\{O_2, O_3\}$. It doesn't matter which one $X$ blocks;
6. $X_3=[0,1]$ or $X_3=[1,1]$
It doesn't matter which $X$ chooses, so long as one segment is left.
7. $O_4=[1,2]$
If $X_3=[0,1]$, then $\{O_1,O_4\}$ and $\{O_2,O_3\}$ are two line segments each with 2 $O$s. If $X_3=[1,1]$, then the two line segments are $\{O_1,O_4\}$ and $\{O_1,O_3\}$. $X$ should probably take the intersection of these two segments since it blocks both. For simplicity, I will assume $X_3=[0,1]$, the the argument works either way;
8. $X_4=[\frac{2}{3}, \frac{4}{3}]$

Now $O$ can take a third point on the top line that has not been blocked. $X$ must then block it since $O$ now has 3 in a row and can win on the next turn if not blocked.
9. $O_5=[2,2]$
10. $X_5=[\frac{1}{2},2]$
At this point, there are two segments, each with two points which are not yet blocked. These are $\{O_1,O_5\}$ and $\{O_2,O_4\}$. $O$ simply takes the intersection of these points.
11. $O_6=[\frac{4}{3},\frac{4}{3}]$

Now $O$ has two line segments, each with 3 points. It doesn't matter which segment $X$ blocks, $O$ will win on the other since $X$ cannot block both.
$N=5$
As we saw previously, we can in 12 moves (six moves by $O$), get an unblocked line with 3 in a row. Lets say we play out this strategy in 3 sections of the real plane, roughly forming a triangle. We will make one move in each section round robin style.
$X$ can either concentrate on eliminating all possibilities in one section, or follow us to each section we move in. If $X$ concentrates on one section, then in two of our sections we will get 3 in a row easily using the above strategy since there is less opposition to our moves.
If $X$ instead chose to follow our round robin strategy in each section, then we would get 3 in a row in all 3 sections using the above strategy. Either way, at least two of them will have 3 in a row.
So long as we didn't make these lines parallel, we can choose a 4th point on the intersection of these lines. This will give us 2 lines of 4 in a row with no blockers. On our next move, we can complete a 5 in a row.
General solution for all $N$
We can simply repeat the $N=5$ strategy for a number of sections on the ever growing real plane. Again, $X$ will be forced to either follow our round robin strategy, or commit to a section. Either way, we will eventually get 4 in a row in at least 2 sections. By placing a point in the intersection of these segments, we will have 2 sets of 5 in a row.
This strategy can be repeated, growing larger and larger each time to achieve all possible $N$.