Let $I\subseteq T$ be a homogeneous ideal. If $I_i=T_i$ for some $i$ then $I_j=T_j,\ \forall j\geq i$.
It is easy to see that it is true for some examples. Does anybody have any ideas for proving it in general cases?
For example, $T=k[x,y]$ and $I=(x,y)$. So we have $T_i=I_i$ for every $i$
I assume you mean $T=k[x_0,\ldots,x_n]$, and $I$ is a homogeneous ideal, and $T_i$ means the degree $i$ part of $T$, namely $T_i=(\mathfrak m)_i$, where $\mathfrak m$ is the ideal $(x_0,\ldots,x_n)$.
Then here's a hint.
Note that $\mathfrak m \cdot \mathfrak m^i = \mathfrak m^{i+1}$.
Thus if $I_i=T_i$ for some $I$, meaning $I_i= \mathfrak m^i/\mathfrak m^{i+1}$ as vector spaces, we get $$ I_{i+1} \supset (\mathfrak m I_i)_{i+1} = (\mathfrak m \cdot \mathfrak m^i)_{i+1} = (\mathfrak m^{i+1})_{i+1}. $$