$k$ is a natural number.
There are three vertices $u,v,w$ such that every two of them doesn't connect to each other.
I think that I need to use ${k+1\choose 2} = \frac{k^2+k}{2} = \frac{k(k+1)}{2}$ in the proof , but actually I don't find any connection beetween this to the sentence.
We can prove it by induction.
if $k=1$, the claim obviously holds for any graph with ${k+1\choose2}=1$ vertices.
Assume for $k-1$ and we'll prove for $k$:
Let $G$ be a triangle-free graph with $k+1\choose2$ vertices.
Let $\varDelta := max\{deg(v) | v\in V(G)\}$.
If $\varDelta \geq k$, then let $v$ be the vertex for which $deg(v) = \varDelta$ and now $S = N(v)$ is an independent set in $G$ with size $\geq k$.
If $\varDelta < k$, pick an arbitrary vertex $v\in V(G)$, remove $\{v\}\cup N(v)$ from $G$. We have removed at most $k$ vertices from the graph, if we have removed less then $k$, remove arbitrary vertices until exactly $k$ vertices have been removed.
In the resulting graph, there are ${k+1\choose2} - k = \frac{k(k+1)}{2} - k = \frac{k(k-1)}{2} = {(k-1)+1\choose2}$ vertices. Hence, by the induction hypothesis, there is an independent set $S$ in this graph such that $|S|=k-1$, and $S\cup\{v\}$ is an independent set in $G$ with size $k$.