The problem is equivalent to: color each edge of a $K_{101}$ by one of 11 colors such that at least two colors are used. Prove that there is a triangle with two sides having the same color, but not the third side. Assume the opposite. Consider 100 edges from a fixed vertex $v_0$. From pigeonhole, there are 10 edges with the same color $c$, so an edge between any two of these 10 vertices must also be color $c$. Thus, we have a $K_{11}$ of color $c$. Let $U$ be the set of these 11 vertices. For any vertex $v \notin U$, if none of the 11 edges joining $v$ with a vertex in $U$ has color $c$, from pigeonhole two of them must have the same color different from $c$, a contradiction. So, at least one of them must have color $c$, which implies all of them must have color $c$. Hence, we can add $v$ to the set $U$, forming a $K_{12}$ of color $c$. We can repeatedly do this until we add all vertices into $U$, so all edges must have color $c$, a contradiction since at least two colors must be used. The bound is sharp. One can construct a counterexample for $K_{100}$.

Consider the obvious graph interpretation, with each edge coloured by one of $11$ colours, i.e. $0,1,\ldots,10$. Let $v$ be a random vertex of the graph, such that $v$ has neighbors of at least two colours. Such a vertex exists, since ohterwise our graph would have edges of only one colour, which is a contradiction. By the Pigeonhole Principle, there is a colour, WLOG let it be colour $c$, such that $v$ has at least $10$ neigbors of this colour. Let $X$ be the set of these neighbors. Note that if there are two vertices in $X$ who are joined with a colour distinct from $c$, we may pick these two vertices and $v$ to finish. Assume from now that all vertices in $X$ are joined by edges coloured in colour $c$. Now, consider another colour $c'$, such that there is at least one edge emanating from vertex $v$ which is of colour $c'$. Let $v'$ be a vertex such that the edge joining $v$ and $v'$ is colour $c'$. Then, for each vertex $v_0 \in X$, if the edge joing $v_0$ and $v'$ is of colour $c$ or $c'$, then we may pick vertices $v,v',v_0$ and we can finish. Otherwise, all edges joining a vertex from $X$ and $v'$ would be of a colour different from $c,c'$. Since there are $9$ remaining colours and $|X| \geq 10$, by the Pigeonhole Principle there are two vertices $v_1,v_2 \in X$, such that the edges joining $v_i$ with $v'$ are of the same colour, distinct from $c,c'$. Thus, we may pick vertices $v_1,v_2$ and $v'$ to finish ($v_1,v_2$ are joined to $v'$ with the same colour, which is distinct from the colour $c$ which joins vertices $v_1,v_2$). In any case, we are done.