Consider a function L from R->R where L(x) = f(x+T) - f(x).
Since f is continuous over R, so is L.

If L(0)=0, we are done and have our antipodal point.
If not, there exists some x for which L(0) < 0 0 > L(T/2)). Therefore for some value of x between 0 and T/2, L(x)=0 which becomes our antipodal point.

Consider a continuous function, P(x) = f(x + T/2) - f(x)
Then,
P(x+T/2) = f(x+T) - f(x+T/2)
So, P(x) + P(x+T/2) = f(x+T) - f(x) = 0
which means P(x+T/2) = -P(x)
If P(x) = 0 for all x, then we are done.
Otherwise, by Intermediate Value Theorem, there exists y such that y > x, y < x+T/2 & P(y) = 0 f(y+T/2) = f(y)

Consider a function L from R->R where L(x) = f(x+T) - f(x).

Since f is continuous over R, so is L.

If L(0)=0, we are done and have our antipodal point.

If not, there exists some x for which L(0) < 0 0 > L(T/2)). Therefore for some value of x between 0 and T/2, L(x)=0 which becomes our antipodal point.

Consider a continuous function, P(x) = f(x + T/2) - f(x)

Then,

P(x+T/2) = f(x+T) - f(x+T/2)

So, P(x) + P(x+T/2) = f(x+T) - f(x) = 0

which means P(x+T/2) = -P(x)

If P(x) = 0 for all x, then we are done.

Otherwise, by Intermediate Value Theorem, there exists y such that y > x, y < x+T/2 & P(y) = 0 f(y+T/2) = f(y)