If you have a strategy that wins more than 50% of the time, your opponent can use the same strategy and also win more than 50% of the time, which is impossible. Therefore, the best possible strategy cannot guarantee winning more than 50% of the time.
But what strategy will guarantee at least 50% against any other strategy?
Part 1: Without the 5 points goal
For now, we will ignore the goal of getting to 5 points, and just try to maximize the value of our score minus the opponent's score. We must find a mixed strategy such that the expected value against any other strategy is at least 0.
Suppose we choose 1 with probability $p_1$, 2 with probability $p_2$, and so on, such that $p_1+p_2+p_3+\ldots=1$.
If the opponent always chooses 1, our expected payoff is $2p_2-p_3-p_4-\ldots=p_1+3p_2-1$. This quantity must be at least 0, or else we are losing. We can repeat this process for each possible number the opponent chooses:
$$-2p_1+2p_3-p_4-p_5-\ldots=-p_1+p_2+3p_3-1\ge0\\
p_1-2p_2+2p_4-p_5-\ldots=2p_1-p_2+p_3+3p_4-1\ge0$$
In general, for all $n\ge2$,
$$2(p_1+p_2+\ldots+p_{n-2})-p_{n-1}+p_n+3p_{n+1}-1\ge0\\
2(p_1+p_2+\ldots+p_{n+1})+p_{n+1}\ge1+3p_{n-1}+p_n$$
Let $S_n=p_1+p_2+\ldots+p_n$. Then
$$3S_{n+1}-S_n\ge1+S_n+2S_{n-1}-3S_{n-2}\\
3S_{n+1}+3S_{n-2}\ge1+2S_n+2S_{n-1}$$
This holds down to $n=1$ if we assume $S_0=S_{-1}=0$. Some linear algebra produces a solution: $$S_1=1/16\\S_2=3/8\\S_3=5/8\\S_4=15/16\\S_5=1$$
So our optimal strategy is: $$p_1=p_5=1/16\\p_2=p_4=5/16\\p_3=1/4$$
This strategy guarantees that in the long run, our points will keep pace with the opponent's points.
Part 2: With the goal
Now we add the goal. This changes things slightly, because if we have four points, it doesn't matter whether we win one point or two, and similarly for the opponent.
4-4
Suppose both players have 4 points. If we get one point or two points, our payoff is 1, and if the opponent gets one point or two points, our payoff is -1. We require our expected payoff to be 0 (any higher is again impossible because of symmetry). Repeating the process with the adjusted payoffs results in the inequality:
$$2S_{n+1}+2S_{n-2}\ge1+S_n+S_{n-1}$$
We can solve this to find:
$$p_1=p_2=p_3=1/3$$
This makes sense, because this situation is basically rock-paper-scissors with the numbers 1, 2, and 3: each number beats one and loses to one. All the numbers greater than 3 are irrelevant because they lose to this strategy.
4-3
This case is asymmetric, which makes it harder to analyze. The side with 4 points wins with one point or two, while the side with 3 points wins if getting two points. We know the situation is equal if the side with 3 points gains one point.
If my methods and calculations were correct, the value of this situation is about $0.239142$ for the player with 4 points (or about 62% chance to win). This number is the real root of a fifth degree polynomial. The strategy for the player with 4 points is
$$p_1\approx0.067558\\p_2\approx0.222986\\p_3\approx0.253375\\p_4\approx0.333977\\p_5\approx0.122105$$
and the strategy for the player with 3 points is the same in reverse:
$$p_1\approx0.122105\\p_2\approx0.333977\\p_3\approx0.253375\\p_4\approx0.222986\\p_5\approx0.067558$$
This result indicates that it is very unlikely that there will be a nice solution for the rest of the problem.
3-3
Using $0.239142$ as the value of winning or losing one point, we can calculate the optimal strategy for a score of 3-3:
$$p_1=p_7\approx0.009228\\p_2=p_6\approx0.191209\\p_3=p_5\approx0.161754\\p_4\approx0.275618$$
I'm not going to continue further than this because the rest of the asymmetric scores will be very tedious to solve.