Theorem. An open set in real sequence space under the ℓ^∞ norm is not totally bounded.

Proof. Consider an open set U containing a point p. Suppose by way of contradiction that U is totally bounded. Then for every ε > 0, there exists a finite ε-net for U. Fix ε, and let m be the number of points in our ε-net, which net we'll denote {S_i}_{i∈{1, ..., m}}. We're going to construct a very special point y, which does not live in U. For all i ∈ {1, ..., m}, we can choose the ith component y_i such that the absolute value of its difference from the ith component of the ith point in the net is strictly greater than ε (that is, |y_i – S_i,i| > ε) but also so that the absolute value of its difference from the ith component of p is less than or equal to ε (that is, |y_i – p_i| ≤ ε). Then for j > m, set y_j = p_j. Then |y – p| ≤ ε, but that means there are points arbitrarily close to p which are not in U, which is an absurd thing to happen to a point in an open set! But that's what I've been trying to tell you this entire time.