Geometrical proof
Geometrical view
Consider the observed sample as a point in n-dimensional Euclidean space and the estimation of the mean as the projection of an observation $x_1,x_2,...,x_n$ onto the model line $x_1=x_2= ... = x_n = \bar{x}$.
The t-score can be expressed as ratio of two distances in this space
- the distance between the projected point and the population mean $$\sqrt{n}(\bar{x}-\mu)$$
- the distance between this point and the observation $$\sqrt{\sum_{i=1}^n(\hat{x}-x_i)^2}$$
This is related to the tangent of the angle between the observation and the line on which it is projected.
$$\frac{t}{\sqrt{n-1} } =\frac{\sqrt{n}(\bar{x}-\mu)}{\sqrt{\sum_{i=1}^n(\hat{x}-x_i)^2}} = \frac{1}{\tan{\theta}}$$

Equivalence t-distribution and angle distribution
In this geometrical view the probability of the t-score being higher than some value is equivalent to the probability of the angle being less than some value:
$$Pr(|T|>t_{n-1,\alpha/2}) = 2 Pr(\theta \leq \theta_{\nu,\alpha}) = \alpha$$
Or
$$\frac{t_{n-1,\alpha/2}}{\sqrt{n-1}} = \frac{1}{\tan \theta_{\nu,\alpha}}$$
You could say that the t-score relates to the angle of the observation with the line of the theoretic model. For points outside the confidence interval (then $\mu$ is further away from $\bar{x}$ and the angle will be smaller) the angle will be below some limit $\theta_{\nu,\alpha}$. This limit will change with more observations. If the limit of this angle $\theta_{\nu,\alpha}$ goes to 90 degrees for large $n$ (the cone shape getting more flat, ie less pointy and long) then this means that the size of the confidence interval becomes smaller and approaches zero.

Angle distribution as relative area of the cap of an n-sphere
Due to symmetry of the joint probability distribution of independent normal distributed variables, every direction is equally probable and the probability for the angle to be within a certain region is equal to the relative area of the cap of an n-sphere.
The relative area of this n-cap is found by integrating the area of a n-frustum:
$$\begin{array}{rcl} 2 Pr(\theta \leq \theta_c)& =& 2 \int_{\frac{1}{\sqrt{1+\tan(\theta_c)^2}}}^1 \frac{(1-x^2)^{\frac{n-3}{2}}}{B(\frac{1}{2},\frac{n-1}{2})} dx \\
&=& \int_{\frac{1}{{1+\tan(\theta_c)^2}}}^1 \frac{t^{-0.5}(1-t)^{\frac{n-3}{2}}}{B(\frac{1}{2},\frac{n-1}{2})} dt \\
&=& I_{\frac{1}{{1+\tan(\theta_c)^2}}}\left(\frac{1}{2},\frac{n-1}{2} \right)
\end{array}$$
where $I_x(\cdot,\cdot)$ is the upper regularized incomplete beta function.
Limit of angle
If $\theta_{n,\alpha}$ goes to 90 degrees for $n \to \infty$ then $t_{n-1,\alpha/2}/\sqrt{n}$ goes to zero.
Or an inverse statement: for any angle smaller than 90 degrees the relative area of that angle on a n-sphere, decreases to zero when $n$ goes to infinity.
Intuitively this means that all area of a n-sphere concentrates to the equator as the dimension $n$ increases to infinity.
Quantitatively we can show this by using the expression
$$\int_{a}^1 \frac{t^{-0.5}(1-t)^{\frac{n-3}{2}}}{B(\frac{1}{2},\frac{n-1}{2})} dt < \int_{a}^1 \frac{(1-a)^{\frac{n-3}{2}}}{B(\frac{1}{2},\frac{n-1}{2})} dt = \frac{(1-a)^{\frac{n-1}{2}}}{B(\frac{1}{2},\frac{n-1}{2})} = L(n)$$
and consider the difference between $L(n+2)$ and $L(n)$.
At some point the decrease in the denominator $$\frac{B(\frac{1}{2},x+1)}{B(\frac{1}{2},x)} = \frac{x}{x+\frac{1}{2}}$$ will be taken over by the decrease in the numerator $$\frac{(1-a)^{\frac{n+1}{2}}}{(1-a)^{\frac{n-1}{2}}} = 1-a$$ and the function $L(n)$ decreases to zero for $n$ to infinity.