Proving $\sum_{i=1}^n(X_i-\overline X_n)^2-\sum_{i=1}^m(X_i-\overline X_m)^2 \sim \chi^2_{n-m}$ 
Suppose $X_1,X_2,\ldots,X_n$ are i.i.d $N(0,1)$ random variables. For $2\le m<n$, let $S_m^2=\sum_{i=1}^m(X_i-\overline X_m)^2$ and $S_n^2=\sum_{i=1}^n(X_i-\overline X_n)^2$ where $\overline X_m=\frac1m\sum_{i=1}^m X_i$ and $\overline X_n=\frac1n\sum_{i=1}^n X_i$. I am trying to prove that $T=S_n^2-S_m^2 \sim \chi^2_{n-m}$.

My idea is to write $T$ as a quadratic form $X^TAX$ where $X=(X_1,\ldots,X_n)^T$ and $A$ is a symmetric matrix of order $n$. Then $T$ would have a $\chi^2$ distribution if and only if $A$ is idempotent, the degrees of freedom of $T$ being the rank of $A$ (or the trace of $A$ since $A$ is idempotent).
Now $S_n^2=X^TA_1X$ where $A_1=I_n-\frac1n \mathbf1_n\mathbf1_n^T$ and $\mathbf1_n$ is a vector of all ones.
If $Y=(X_1,\ldots,X_m)^T$, then similarly, $S_m^2=Y^TA_2Y$ with $A_2=I_m-\frac1m \mathbf1_m\mathbf1_m^T$.
So I think $$T=X^TA_1X-Y^TA_2Y=X^TAX\,,$$
where $$A=A_1-\begin{pmatrix}A_2 & O_{m\times \overline{n-m}} \\ O_{\overline{n-m}\times m} & O_{n-m}\end{pmatrix}$$
I can show that $A_1$ and $A_2$ are idempotent, but verifying $A$ is idempotent is somewhat cumbersome. Is there any easier way out? Alternatively, is $S_n^2-S_m^2$ independent of $S_m^2$? I understand this would solve the problem since $S_n^2 \sim \chi^2_{n-1}$ and $S_m^2 \sim \chi^2_{m-1}$. Again, according to a theorem on quadratic forms, I just need to show $T$ is non-negative definite. Then from $S_n^2 \sim \chi^2_{n-1}$ and $S_m^2 \sim \chi^2_{m-1}$, it would follow that $T\sim \chi^2_{(n-1)-(m-1)}$. Any suggestions are welcome.

I think your approach is good.
I can provide a (not to tedious) proof that the matrix $A$ is idempotent of order 2.
Using your notations, we have $$A_1 = I_n - \frac{1}{n} \boldsymbol{1}_n \boldsymbol{1}^T_n$$
and I'll change a tiny bit your notation for $A_2$ :
$$A_2 = I_m^* - \frac{1}{m} \boldsymbol{1}_m^* \,\,{\boldsymbol{1}^{*}_m}^T$$
where $I_m^* \in \mathbb{R}^{n\times n}$ is a diagonal matrix with $m$ ones and then $n - m$ zeros on its diagonal, and $\boldsymbol{1}_m^*\in \mathbb{R}^n$ is a vector of $m$ ones then $n - m$ zeros.
So $S^2_n = X^T A_1 X$ and $S^2_m = X^T A_2 X$ and thus: $$T = S^2_n - S^2_m = X^T (A_1 - A_2) X = X^T A X $$
with $A = A_1 - A_2$.
Let's show that $A^2 = A$.
We have that :
$$A^2 = (A_1 -A_2)^2 = A_1^2 + A_2^2 -A_1A_2 - A_2 A_1 = A_1 + A_2 - A_1A_2 - A_2 A_1$$
using the fact that $A_1 ^2 = A_1$ and $A_2^2 = A_2$.
Now we can show that $A_1A_2 = A_2A_1 = A_2$.
$$
\begin{array}{ll}
A_1 A_2 & = (I_n - \frac{1}{n} \boldsymbol{1}_n \boldsymbol{1}^T_n)(I_m^* - \frac{1}{m} \boldsymbol{1}_m^* \,\,{\boldsymbol{1}^{*}_m}^T)\\
 & = A_2 -\frac{1}{n}\boldsymbol{1}_n \boldsymbol{1}^T_n I_m^* + \frac{1}{n m }\boldsymbol{1}_n \boldsymbol{1}^T_n \boldsymbol{1}_m^* \,\,{\boldsymbol{1}^{*}_m}^T \\
 &= A_2 -\frac{1}{n}\boldsymbol{1}_n {\boldsymbol{1}^T_m}^* + \frac{1}{n}\boldsymbol{1}_n \,{\boldsymbol{1}^{*}_m}^T
\end{array}
$$
by noticing that $\boldsymbol{1}^T_n \boldsymbol{1}_m^* = m$ and that $\boldsymbol{1}_n^T I_m^* = {\boldsymbol{1}_m^*}^T$.
Therefore $A_1 A_2 = A_2 . $
An almost identical computation gives that $A_2A_1 = A_2$.
So in the end $(A_1 - A_2)^2 = A_1 + A_2 - A_2 - A_2 = A_1 - A_2$.
So to conclude on the distribution on $T$, we have that $T \sim \chi^2_{rank(A)}$.
As $A^2 = A$, $rank(A) = trace(A) = trace(A_1) - trace(A_2) = (n-1) - (m-1) = n - m$.
Hence: $$T \sim \chi^2_{n - m}.$$
