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Let's imagine a game like tic-tac-toe, but you have to have 3 of the same sign in the same line or colon... So the diagonals don't count!

Then I have to prove that the second player to play can always be sure not to lose... But how do I do that?

I am searching a simple argument, not just like constructing a game tree...

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    How much of a game tree (up to the many symmetries of the board in this variant) have you constructed before concluding that is not a feasible way to proceed? – 2017-02-22
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    I am searching a simpler argument, because constructing a game tree is long and isn't a nice proof... – 2017-02-22
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    Duplicate of http://math.stackexchange.com/q/2156626/265466 – 2017-02-22

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Suppose X plays first. The only thing O needs to do to avoid losing is to manage to have one mark in each row and one mark in each column. It's easy to find a strategy that can achieve that.

For example, place the first two Os in the same column and the same row as X's first play. Before O's next move, the board looks like this, up to symmetries

X O .
O . .
. . .

plus two Xs that the first player has put down in the mean time. If the bottom right corner is still free, O plays that and has now forced a draw. Otherwise the situation (still up to symmetry) is one of

X O X     X O .     X O .
O . .     O . X     O X .
. . X     . . X     . . X

In each of these cases, O can now play in the last column, and X is then powerless to prevent O from occupying the bottom row the next time, again forcing a draw.

Of course, if you already know that ordinary tic-tac-toe always ends in a draw with perfect play, you don't even need this much analysis: Second player simply pretends he's playing ordinary tic-tac-toe. Since this is enough to prevent the first player from winning, it will also prevent the first player from meeting the strictly harder victory condition of the variant game.