@Kiwiakos gave you the intuition. Here is the corresponding analysis that you asked for. The European plain vanilla call delta is given by
\begin{equation} \frac{\partial C_0}{\partial S_0} = \mathcal{N} \left( d_+ \right), \end{equation}
where
\begin{equation} d_+ = \frac{1}{\sigma \sqrt{T}} \left( \ln \left( \frac{S_0}{K} \right) + \left( r - \frac{1}{2} \sigma^2 \right) T \right) \end{equation}
Differentiating again w.r.t. $T$ yields
\begin{eqnarray} \frac{\partial^2 C_0}{\partial S_0 \partial T} & = & \mathcal{N}' \left( d_+ \right) \frac{\partial d_+}{\partial T}\\ & = & \mathcal{N}' \left( d_+ \right) \frac{1}{2 \sigma T \sqrt{T}} \left( -\ln \left( \frac{S_0}{K} \right) + \left( r - \frac{1}{2} \sigma^2 \right) T \right). \end{eqnarray}
This term is negative (delta is increasing as the time-to-maturity becomes shorter) if
\begin{equation} -\ln \left( \frac{S_0}{K} \right) + \left( r - \frac{1}{2} \sigma^2 \right) T < 0 \qquad \Leftrightarrow \qquad S_0 > K \exp \left\{ \left( r - \frac{1}{2} \sigma^2 \right) T \right\} := S^*(T) \end{equation}
In the limit we have
\begin{equation} \lim_{T \downarrow 0} S^*(T) = K \end{equation}
as expected.