François Brunault commented that the maker can get two moves to start on some hexagonal lattice (a lattice generated by unit vectors with an angle of $60$ degrees). In fact, by the fourth move, the maker can get three moves to start in a hexagonal lattice, and can choose these to be the vertices of an equilateral triangle of size one, so that the breaker has not played in this lattice yet.

Proof: Let the maker's first play be the origin $A$. Rotate the coordinates after the breaker's first move to that this play be ignored. Let the maker's second move be $B = (2x,0)$ where $x$ is a transcendental smaller than $1$. There are $4$ hexagonal lattices $H_{A,C}, H_{A,D}, H_{B,C}, H_{B,D}$ containing one element of $\lbrace (0,0), (2x,0)\rbrace$ and one element of $\lbrace C=(x,\sqrt{1-x^2}),D=(x,-\sqrt{1-x^2}) \rbrace$, the points of distance $1$ from the maker's first two moves. These hexagonal lattices have the property that the breaker hasn't played in any of them yet, and their pairwise intersections are empty or a point in $\lbrace A,B,C,D\rbrace.$ So, either the second play of the breaker misses both $H_{A,C}$ and $H_{B,C}$ or both $H_{A,D}$ and $H_{B,D}$. By symmetry, we can assume the second play of the breaker misses $H_{A,C}$ and $H_{B,C}$. Let the third play of the maker be $C$. This gives the maker and unopposed pair of points in both $H_{A,C}$ and $H_{B,C}$. The third play of the breaker can only be in one of these lattices. On the fourth move of the maker, the maker can play in the other to make an unopposed equilateral triangle of side length $1$. $\blacksquare$

Next, even if the maker is constrained to play in this lattice, the maker can force $5$ in a row by move $11$. Note that if the maker has an "open $3$" of $3$ points in row with two open spaces on either side, $- - \circ \circ \circ - -$, then the breaker has to respond immediately either just to one side or the other, or else the maker can make $5$ in a row in $2$ move moves. To avoid an explosion of cases, we'll let the breaker play both sides of an open $3$. Perhaps without this, the maker could force $5$ in a row in fewer moves.

We'll show that whatever the breaker's fourth move is, the maker can still force $5$ in a row quickly. By symmetry, we can assume the breaker's fourth move is in a sixth of the lattice.

o o . x x x x x
 o . x x x x x
. . . x x x x x
 . . . . x x x

5 o o . x x x x x
   o . x x x x x
  . . . x x x x x
   . . . . x x x

This play dies not technically make an open $3$ since only one space to the right is open. So, the breaker does not have to respond to either side immediately. The other possibility is $2$ to the left. We will let the breaker play in all $3$ positions simultaneously.

x x 5 o o x x x x x x
       o . x x x x x
      . . . x x x x x
       . . . . x x x

         x
x x 5 o o x x x x x x
       o . x x x x x
      6 . . x x x x x
     x . . . . x x x

   x     x
x x 5 o o x x x x x x
     7 o . x x x x x
      6 . . x x x x x
     x x . . . x x x

   x   x x
x x 5 o o x x x x x x
     7 o . x x x x x
    8 6 . . x x x x x
   x x x . . . x x x

   x   x x
x x 5 o o x x x x x x
     7 o . x x x x x
    8 6 9 . x x x x x
   x x x . . . x x x

François Brunault commented that the maker can get two moves to start on some hexagonal lattice (a lattice generated by unit vectors with an angle of $60$ degrees). In fact, by the fourth move, the maker can get three moves to start in a hexagonal lattice, and can choose these to be the vertices of an equilateral triangle of side $1$ so that the breaker has not played in this lattice yet.

Proof: Let the maker's first play be the origin $A$. Rotate the coordinates after the breaker's first move so that this play be ignored. Let the maker's second move be $B = (2x,0)$ where $x$ is a transcendental smaller than $1$. There are $4$ hexagonal lattices $H_{A,C}, H_{A,D}, H_{B,C}, H_{B,D}$ containing one element of $\lbrace (0,0), (2x,0)\rbrace$ and one element of $\lbrace C=(x,\sqrt{1-x^2}),D=(x,-\sqrt{1-x^2}) \rbrace$, the points of distance $1$ from the maker's first two moves. These hexagonal lattices have the property that the breaker hasn't played in any of them yet, and their pairwise intersections are empty or a point in $\lbrace A,B,C,D\rbrace.$ So, either the second play of the breaker misses both $H_{A,C}$ and $H_{B,C}$ or both $H_{A,D}$ and $H_{B,D}$. By symmetry, we can assume the second play of the breaker misses $H_{A,C}$ and $H_{B,C}$. Let the third play of the maker be $C$. This gives the maker an unopposed pair of points in both $H_{A,C}$ and $H_{B,C}$. The third play of the breaker can only be in one of these lattices. On the fourth move of the maker, the maker can play in the other to make an unopposed equilateral triangle of side length $1$. $\blacksquare$

Next, even if the maker is constrained to play in this lattice, the maker can force $5$ in a row by move $11$. Note that if the maker has an "open $3$" of $3$ points in row with two open spaces on either side, $- - \circ \circ \circ - -$, then the breaker has to respond immediately either just to one side or the other, or else the maker can make $5$ in a row in $2$ more moves. To avoid an explosion of cases, we'll let the breaker play both sides of an open $3$. Perhaps without this, the maker could force $5$ in a row in fewer moves.

We'll show that whatever the breaker's fourth move is, the maker can still force $5$ in a row by the eleventh move. By symmetry, we can assume the breaker's fourth move is in a sixth of the lattice between two of the perpendicular bisectors of the triangle's sides. We'll consider two possible moves within this sixth individually, and then all others.

     x x x x x
o o . x x x x x
 o . x x x x x
. . . x x x x x
 . . . . x x x 

       x x x x x         
5 o o . x x x x x
   o . x x x x x
  . . . x x x x x
   . . . . x x x

This play technically does not make an open $3$ since only one space to the right is open. So, the breaker does not have to respond to either side immediately. The other possibility is $2$ to the left. We will let the breaker play in all $3$ positions simultaneously.

           x x x x x
x x 5 o o x x x x x x
       o . x x x x x
      . . . x x x x x
       . . . . x x x

         x x x x x x
x x 5 o o x x x x x x
       o . x x x x x
      6 . . x x x x x
     x . . . . x x x

   x     x x x x x x
x x 5 o o x x x x x x
     7 o . x x x x x
      6 . . x x x x x
     x x . . . x x x

   x   x x x x x x x
x x 5 o o x x x x x x
     7 o . x x x x x
    8 6 . . x x x x x
   x x x . . . x x x

   x   x x x x x x x
x x 5 o o x x x x x x
     7 o . x x x x x
    8 6 9 . x x x x x
   x x x . . . x x x

In the next sequence, I'll show the breaker's response immediately, again letting the breaker block both sides of open threes.

Again, the maker's 9th move creates $2$ open threes, through the $7$ and through the $5$ and $8$, so with that choice of $4$th move by the breaker, the maker can construct $5$ in a row by move $11$.

Next we let the breaker's 4th move block every other possibility in that sixth of the lattice, which will cover all of those possibilities simultaneously.

This play dies not technically make an open 3 since only one space to the right is open. So, the breaker does not have to respond to either side immediately. The other possibility is two to the left. So, we will let the breaker play in all 3 positions simultaneously.

Again, the $9$th move creates 2 open threes, so the maker can construct $5$ in a row by move $11$.

In the next sequence, I'll show the breaker's response immediately, again letting the breaker block both sides of open $3$s.

Again, the maker's 9th move creates two open $3$s, through the $7$ and through the $5$ and $8$, so with that choice of $4$th move by the breaker, the maker can construct $5$ in a row by move $11$.

Next we let the breaker's $4$th move block every other possibility in that sixth of the lattice, which will cover all of those possibilities simultaneously.

This play dies not technically make an open $3$ since only one space to the right is open. So, the breaker does not have to respond to either side immediately. The other possibility is $2$ to the left. We will let the breaker play in all $3$ positions simultaneously.

Again, the $9$th move creates two open $3$s, so the maker can construct $5$ in a row by move $11$.