Yunone is right that this fact is basically equivalent to Pasch's postulate, which can't be proved from Euclid's axioms. There are many such gaps in the axioms corresponding to incidence results that are obviously true if you draw a picture and hence were left unstated, like that the two circles in his first proposition intersect, and that is why Euclid's axioms are said to be insufficient.
So, one answer to your question is that the result in question is true as a consequence of Pasch's postulate if you are using an axiom system that includes it. But there is no geometric axiom system that can be considered the standard formulation of Euclidean geometry today. Rather, the standard formulation nowadays is in terms of the Cartesian plane. Points are ordered pairs, lines are solution sets of linear equations, distance is given by the usual formula, etc.
In this formulation Pasch's postulate is a theorem that can be proven.
A straightforward way to see it is as a consequence of Menelaus' theorem, which can be proven purely algebraically. If a line intersected the interiors of the three sides, this theorem would give us the product of three positive numbers as -1.
But there's a more direct though ugly way to see it. Given a line intersecting the interiors of all sides of a triangle, you can use coordinates to transform this statement into a series of equations and inequalities between real numbers and show they cannot be satisfied. Here are the unattractive details:
Translations preserve lines and betweenness, so translate one vertex to the origin. Linear transformations also preserve lines and betweennes, so transform the two other vertices to (0,1) and (1,0). So, without loss of generality we can assume our triangle has vertices (0,0), (0,1), and (1,0). Now, suppose a line intersects all three sides. If it were vertical, it would be parallel to one side and thus couldn't intersect all three. Thus, the line is the graph of a function f(x). Suppose it intersects the interior of the two edges meeting the origin. Then
$0 < f(0) < 1$ and there is an r in $(0,1)$ such that $f(r) = 0$. Thus $f(x)$ is decreasing so it can be written as $f(x) = mx + b$ where$ m < 0$ and $0 < b < 1$, and also $f(1)=m+b<f(r)=0$ from the fact it is decreasing.
$1-x$ is the equation of the third side. Let's look at the intersection between this side and the line defined by $f(x)$. If $1-x = mx+b$, then $(m+1)x=1-b$. If it intersects the interior of this side, $0 < x < 1$. Thus $m+1 > 0$, so $1-b<m+1$, so $m+b <0$. But $m+b > 0$ was proved above. So it's impossible for any line to intersect the interiors of all three sides.
This proof is ugly, but it demonstrates that Pasch's postulate is a theorem that you can prove in the Cartesian model of Euclidean geometry.
I think I can clear up some misunderstanding. A line contains more than just two points. A line is made up of infinitely many points. It is however true that a line is determined by 2 points, namely just extend the line segment connecting those two points.
Similarly a plane is determined by 3 non-co-linear points. In this case the three points are a point from each line and the point of intersection. We are not creating a new point when the lines intersect, the point was already there.
This is not the same thing as saying that there are 5 points because there are two from each line and the point from their intersection.
Best Answer
If you pick two points on a plane and connect them with a straight line then every point on the line will be on the plane.
Given two points there is only one line passing those points.
Thus if two points of a line intersect a plane then all points of the line are on the plane.