The easiest approach is to calculate the probability that you get no red marble and subtract that from $1$.
In order to get no red marble, you must (a) get a non-red marble with your first draw and then (b) get another non-red marble on your second draw. The probability of (a) is $\frac5{10}=\frac12$, since there are $2+3=5$ non-red marbles. Similarly, if (a) occurred, the probability of (b) is $\frac49$, since there are now only nine marbles in the bag, of which four are non-red. The combined probability of (a) and (b) occurring is therefore $\frac12\cdot\frac49=\frac29\;,$ and the probability of getting at least one red marble is $1-\frac29=\frac79\;,$ just as you calculated.
Of course you could work the problem directly, by calculating the probability of getting exactly one red marble and the probability of getting two red marbles and adding these probabilities. If you do it directly, you have $\frac12\cdot\frac49=\frac29$ as the probability of getting two red marbles; $\frac12\cdot\frac59=\frac5{18}$ as the probability of getting a red followed by a non-red marble; and $\frac12\cdot\frac59=\frac5{18}$ as the probability of getting a non-red followed by a red marble, for a total probability of $\frac29+\frac5{18}+\frac5{18}=\frac79$ of getting at least one red marble.
Clearly, though, it’s easier to consider the one ‘bad’ case of getting no red marbles than to have to work with the several ‘good’ cases. Had you been drawing three marbles instead of four, and had the problem asked for the probability of getting at least two red marbles, it would have been just about a toss-up, since there are two cases either way: $0$ or $1$ red marble (‘bad’) versus $2$ or $3$ red marbles (‘good’).