Let $f$ be a continuous $2\pi$-periodic function on $\mathbb{R}$. I'm trying to show that \begin{align}\tag{1} \lim_{T \rightarrow \infty} \frac{1}{2T}\int_{-T}^{T} f(x)e^{-ix\xi}dx = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{-ix\xi}dx \end{align} for all $\xi \in \mathbb{R}$.
It's easy to show that if $T = k\pi$ for some $k \in \mathbb{Z}$ and if $\xi \in \mathbb{Z}$, then $$ \frac{1}{2T}\int_{-T}^{T} f(x)e^{-ix\xi}dx = \frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)e^{-ix\xi}dx. $$ This implies that (1) holds for $\xi \in \mathbb{Z}$ whenever the left-hand side of (1) exists.
How can I show (1) in general?
What you are trying to show does not seem to be correct. As a counterexample let us consider $f(x) = e^{ikx}$, where $k \in \Bbb{Z}$. Then
\begin{eqnarray*} \frac{1}{2T}\int_{-T}^{T}e^{ikx}\cdot e^{-ix\xi}\, dx & = & \frac{1}{2T}\cdot\begin{cases} 2T, & k=\xi\\ \frac{e^{i\left(k-\xi\right)x}}{i\left(k-\xi\right)}\bigg|_{x=-T}^{T}, & k\neq\xi \end{cases}\\ & = & \begin{cases} 1, & k=\xi\\ \frac{e^{i\left(k-\xi\right)T}-e^{-i\left(k-\xi\right)T}}{2Ti\left(k-\xi\right)}, & k\neq\xi \end{cases}\\ & = & \begin{cases} 1, & k=\xi\\ \frac{2i\cdot\sin\left(\left(k-\xi\right)T\right)}{2Ti\left(k-\xi\right)}, & k\neq\xi \end{cases}\\ & = & \begin{cases} 1, & k=\xi\\ \frac{\sin\left(\left(k-\xi\right)T\right)}{T\left(k-\xi\right)}, & k\neq\xi \end{cases}\\ & \xrightarrow[T\rightarrow\infty]{} & \begin{cases} 1, & k=\xi\\ 0, & k\neq\xi \end{cases} \end{eqnarray*}
But if we plug in $T= \pi$ in the above calculation, we get \begin{eqnarray*} \frac{1}{2T}\int_{-T}^{T}e^{ikx}\cdot e^{-ix\xi}\, dx & = & \begin{cases} 1, & k=\xi\\ \frac{\sin\left(\pi\left(k-\xi\right)\right)}{\pi\left(k-\xi\right)}, & k\neq\xi \end{cases}\\ & = & \begin{cases} 1, & k=\xi\\ -\frac{\sin\left(\pi\xi\right)}{\pi\left(k-\xi\right)}, & k\neq\xi \end{cases} \end{eqnarray*} which is (for $\xi \notin \Bbb{Z}$) not the same as the result above.