@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.