Every graph on $6$ vertices with $\alpha(G) < 3$ contains at least $2$ copies of $K_3$ I have to show that every graph on $6$ vertices with $\alpha(G) < 3$ contains at least $2$ copies of $K_3$.

$α(G)$ is the largest number of independent vertices and $ω(G)$ is the clique number.
$ω(G) >3$ is clear, this means that we have $K_4$ which contains at least $2$ copies of $K_3$
How do I do $\omega(G) = 3$? Am I somehow supposed to use $R(3,3) = 6$?
 A: Here is an elementary proof but somewhat involved proof. We will repeatedly use the reasoning that for any three vertices we must have edge as otherwise they form an independent set of size 3.
Let $v_1, \ldots, v_6$. Let us pick $v_1$. Then we have a few cases.
Case 1: Suppose the degree of $v_1 \ge 4$.
Without loss of generality that $v_1$ is adjacent to $v_2, v_3, v_4, v_5$. Then there is an edge amongst the three vertices $v_2, v_3, v_4$. Again without loss of generality we can assume that $v_2, v_3$ are connected. Then we again must have an edge between $v_2, v_4, v_5$. Thus have two $K_3$.
Case 2: Suppose the degree of $v_1$ is 3.
Then assume these are $v_2, v_3, v_4$ and we must have an edge between these three and that forms our first $K_3$. Assume this edge is $v_2, v_3$. Assume that $v_3, v_4$ and $v_2, v_4$ are not connected. Otherwise we have our second $K_3$
Now since $v_1$ is not adjacent to $v_5, v_6$ we must have that $v_5, v_6$ are connected. Now similarly there must be an edge amongst $v_3, v_4, v_5$. WLOG there is an edge between $v_3$ and $v_5$. Now look at $v_2, v_4, v_5$. Now if $v_2$ is connected to $v_5$. Then $v_2, v_3, v_5$ is our second $K_3$. Otherwise we have $v_4$ is connected to $v_5$. Now if we look at $v_3, v_4, v_6$. Whatever $v_6$ is connected to along with $v_5,v_6$ forms our second $K_3$.
Case 3: Suppose the degree of $v_1$ is at most 2. Then pick a vertex with degree greater than 2 and repeat the above. If all vertices have degree less than 2. Then unless $G$ is two disjoint copies of $K_3$ this is not possible.
To see suppose all vertices have degree 2. Then $G$ is either two disjoint copies of $K_3$ or is $C_6$ and $\alpha(C_6) = 3$. Now if we have a vertex of degree 1, assume it is $v_1$, and assume that $_1$ is connected to $v_2$. Then we must have that $v_3, v_4, v_5, v_6$ must form a $K_4$ as otherwise $v_1$ and two vertices that are not connected form an independent set of size 3. But we know the max degree is at most 2.
Thus in all cases we two $K_3$s.
A: 
Am I somehow supposed to use $R(3,3)=6$?

We can use this.  If we replace "blue edge" with "edge" and "red edge" with "non-edge" in the usual definition of $R(r,s)$, we get:

there exists a least positive integer $R(r,s)$ for which every graph on $R(r, s)$ vertices contains a clique on $r$ vertices or an independent set on $s$ vertices

$R(3,3)=6$ implies every graph on $6$ vertices satisfying $\alpha(G)<3$ has a $3$-vertex clique.  Let's say the $3$ vertices $\{a,b,c\}$ form a clique and the $3$ other vertices are $\{x,y,z\}$.  Thus far we know these edges exist:

To get a second triangle, we keep going:

*

*If $z$ (and more generally any vertex in $\{x,y,z\}$) is adjacent two vertices in $\{a,b,c\}$, we have a second triangle such as in

So assume every vertex in $\{x,y,z\}$ has at most one neighbor in $\{a,b,c\}$.


*If $\{x,y,z\}$ is a triangle, we are done, so assume two of $\{x,y,z\}$ are non adjacent; by symmetry, assume $y$ and $z$ are not adjacent.



*By the first bullet point, $y$ is adjacent to at most one of $\{a,b,c\}$, and $z$ is adjacent to at most one of $\{a,b,c\}$, which implies there is a vertex in $\{a,b,c\}$ which is neither adjacent to $y$ nor $z$; by symmetry suppose it's $a$:

However, $\{a,y,z\}$ form an independent set of size $3$, violating $\alpha(G)<3$.
In summary: either (a) $\{x,y,z\}$ is a second $K_3$, (b) one of the vertices in $\{x,y,z\}$ have two neighbors in $\{a,b,c\}$ giving a second $K_3$, or (c) we violate $\alpha(G)<3$ by finding an independent set of size $3$.
A: Consider your graph $G=(V,E)$ as a subgraph of the complete graph $K_6$. Color each edge of $K_6$ blue if it's an edge of $G$, red otherwise. You want to show that, if this $2$-colored $K_6$ contains no red triangle, then it must contain (at least) two blue triangles. To put it more symmetrically, you want to show that there must be at least two monochromatic (all red or all blue) triangles in $K_6$.
Note that a bichromatic ($2$-colored, not monochromatic) triangle has exactly two bichromatic angles. (A bichromatic angle consists of a blue edge and a red edge with a common vertex.)
How many bichromatic angles can there be? The number of bichromatic angles at a vertex $v$ is equal to $br$ where $b$ is the number of blue edges and $r$ the number of red edges incident with $v$. Since $b+r=5$ we have $br\le6$, that is, at most $6$ bichromatic angles at each vertex, so at most $36$ bichromatic angles all told.
Since each bichromatic triangle contains two bichromatic angles, at most $18$ of the $\binom63=20$ triangles in $K_6$ are bichromatic, whence at least two of them are monochromatic.
Generalizing this argument leads to A. W. Goodman's formula for the minimum possible number of monochromatic triangles in a $2$-coloring of the edges of $K_n$.
