Is there a simple direct proof of the Open Mapping Theorem from the Uniform Boundedness Theorem? The Open Mapping Theorem, the Bounded Inverse Theorem, and the Closed Graph Theorem are equivalent theorems in that any can be easily obtained from any other.  The Closed Graph Theorem also easily implies the Uniform Boundedness Theorem.  But is there a simple way to obtain any of the other three results from Uniform Boundedness, or is Uniform Boundedness really a "lower-level" result than the others?
 A: I don't know whether you'll consider this "simple", but here is a proof.  I distilled it from Eric Schechter's Handbook of Analysis and its Foundations, which has a proof of a more general statement at 27.35.  The last part is from Folland's Real Analysis, Theorem 5.10.  
Suppose $X,Y$ are Banach spaces and $T : X \to Y$ is surjective.  We wish to show $T$ is open.  Let $B$ be the open unit ball of $X$; it suffices to show $T(B)$ contains a neighborhood of $0 \in Y$.
The first step is to show that the closure $\overline{T(B)}$ contains a neighborhood of 0.  The usual method is to use the Baire category theorem: if not, then $Y = \bigcup_{n=1}^\infty n \overline{T(B)}$ meaning that $Y$ is meager.  We will use the uniform boundedness principle instead.
For each $n$, construct a new norm $\|\cdot\|_n$ on $Y$ defined by
$$\|y\|_n := \inf\{\|u\|_X+n\|v\|_Y : u \in X, v \in Y, v+Tu=y\}.\tag{*}$$
It is straightforward to verify this is a norm.  Now let $Z$ be a countable direct sum of copies of $Y$, i.e., $Z$ is the vector space of all finitely supported functions $f : \mathbb{N} \to Y$, with the pointwise addition and scalar multiplication.  Equip $Z$ with the norm
$$\|f\|_Z := \sup_n \|f(n)\|_n.$$
Then for each $n$, define a linear operator $S_n : Y \to Z$ by $(S_n y)(n) = y$ and $(S_n y)(k) = 0$ for $k \ne n$.  Note that $\|S_n y\|_Z = \|y\|_n$.  
Now by taking $u=0$, $v=y$ in (*), we obtain $\|y\|_n \le n \|y\|_Y$, so each $S_n$ is bounded.   Moreover, by the surjectivity of $T$, for each $y \in Y$ there exists $x \in X$ with $Tx=y$.  Taking $u=x$ and $v=0$ in (*) we see that $\| y\|_n \le \|x\|_X$ independent of $n$; hence $\{S_n\}$ is pointwise bounded.  By the uniform boundedness theorem, there is a constant $C < \infty$ such that $\|S_n\|_{Y \to Z} \le C$ for all $n$.
Fix $\delta < 1/C$.  I claim that $\overline{T(B)}$ contains a ball of radius $\delta$ centered at 0.  For suppose $\|y\|_{Y} < \delta$; then $$\|y\|_n = \|S_n y\|_Z \le \|S_n\|_{Y \to Z} \|y\|_Y \le C \delta < 1$$ for every $n$.  Hence for each $n$ there exists $u_n \in X$, $v_n \in Y$ with $y = v_n + T u_n$ and $\|u_n\|_X + n\|v_n\|_Y < 1$.  In particular, $\|v_n\|_Y < 1/n$, so we have $T u_n \to y$ where $u_n \in B$.  Thus $y \in \overline{T(B)}$ and the proof of this step is complete.
The rest of the proof proceeds as usual.  We can show $T(B)$ contains a ball of radius $\delta/2$ centered at 0.   Suppose $\|y\|_Y < \delta/2$, so that by the first step and scaling, $y \in \overline{T(B_{1/2})}$.  Hence there is $x_1$ with $\|x_1\|_{X} < \frac{1}{2}$ and $\|y - Tx_1\|_Y < \delta/4$.  Repeating this process inductively, we construct $x_n$ with $\|x_n\|_X < 2^{-n}$ and $\left\| y - \sum_{k=1}^n Tx_k\right\|_Y < \delta 2^{-(n+1)}$.  Thus $\sum_{k=1}^\infty T x_k$ converges in $Y$ to $y$.  Summing a geometric series, $\sum_{k=1}^\infty \|x_k\| < 1$, so by completeness of $X$, $\sum_{k=1}^\infty x_k$ converges in $X$ to some $x$ with $\|x\|_X < 1$.  And by continuity of $T$, $Tx=y$.  So we have shown $y \in T(B)$.
Note that this proof works even if $Y$ is not Banach, so long as the uniform boundedness principle holds on $Y$.  This happens iff $Y$ is barreled, and indeed the proof in Schecheter shows that both uniform boundedness and open mapping are equivalent to being barreled.
