Check my proof that continuous linear functionals are bounded I am trying to prove the following claim:

Proposition: Suppose $T$ is a linear functional from $(V, \|\cdot\|_V)$ to $(W, \|\cdot\|_W)$, normed vector spaces. If $T$ is continuous at any point $x \in V$, then it is bounded everywhere, that is: there exists some absolute constant $C > 0$ such that
  $$
\|Tu\|_W \leq C \|u\|_V \qquad \text{for all $u \in V$}. 
$$

Proof: It suffices to show that $\|T\| < \infty$. Suppose for contradiction that it is not finite, equivalently, that $\sup_{\|u\| = 1} \|Tu\|_W= + \infty$. Then, this means that for every $R > 0$, there is some non-zero point $u$ such that $\|Tu\|_W > R$. But now notice using continuity and also linearity, we have that for each $\epsilon > 0$, there is some $\delta(\epsilon)$ such that $\|u\|_V \leq \delta(\epsilon)$ implies 
$\|Tu\|_W \leq \epsilon$. So set $R = \epsilon/\delta(\epsilon)$, and oberve there is a point $u$ with norm 1 such that $\delta(\epsilon)\|Tu\|_W > \epsilon$. But now let $y = \delta(\epsilon) u$, and since the norm is homogeneous, it follows that $\|Ty\|_w > \epsilon$, even though $\|y\| \leq \delta(\epsilon)$, a contradiction. Thus $\|T\| <\infty$, and hence $T$ is bounded.  

Question: Can you comment on the correctness of my proof above?
 A: Our OP Drew Brady's proof seems essentially correct to me, and he does mention that linearity is used to affirm continuity at $0$, although it probably wouldn't have hurt if he had spent a few more words explaining just how this works.  The only other remark I have on the question proper is that the term linear functional is generally used to refer to a linear map from a vector space to its field of scalars, and here it is used for a transformation $T:V \to W$ 'twixt two normed spaces, which would I believe conventionally be called a linear map or linear function.  Not really too big of a deal, I suppose, but for the sake of clarity I point out this terminological distinction.
Having said these things, I would prove the assertion in question as follows;
note the following line does not use reductio ad absurdum, that is, proof by contradiction:
Let $x \in V$ be a point at which $T:V \to W$ is continuous; then given any real $\epsilon > 0$, there exists real $\delta > 0$ such that
$\Vert y - x \Vert_V < \delta \Longrightarrow \Vert T(y) - T(x) \Vert_W < \epsilon; \tag 1$
now if we take any
$z \in V, \; \Vert z \Vert_V < \delta, \tag 2$
we can set
$y = x + z, \tag 3$
whence
$z = y - x; \tag 4$
thus
$\Vert y - x \Vert_V = \Vert z \Vert_V < \delta \Longrightarrow \Vert T(y) - T(x) \Vert_W < \epsilon, \tag 5$
and since
$T(y) - T(x) = T(y - x) = T(z), \tag 6$
we may write (5) as
$\Vert z \Vert_V < \delta \Longrightarrow \Vert T(z) \Vert_W < \epsilon, \tag 7$
which is precisely affirms that $T$ is continuous at $0$.  It is easy to see that the argument ensconced in (1)-(7) may in fact be walked backwards to show that continuity at $0$ implies continuity at any $x \in V$.
From (7), we may deduce the boundedness of $T$:  let
$v \in V, \; \Vert v \Vert_V = 1, \tag 8$
and let $r \in \Bbb R$ with $0 < r < 1$; then 
$\Vert r \delta v \Vert_V = r \delta \Vert v \Vert_V < \delta, \tag 9$
whence (7) implies
$r \delta \Vert T(v) \Vert_W  = \Vert r \delta T(v) \Vert_W = \Vert T(r \delta v) \Vert_W  < \epsilon = \epsilon \Vert v \Vert_V, \tag{10}$
whence
$\Vert T(v) \Vert_W < \dfrac{\epsilon}{r \delta} \Vert v \Vert_V; \tag{11}$
(11) binds for every $v \in V$ with $\Vert v \Vert_V = 1$; for any $0 \ne u \in V$ we may write
$u = \Vert u \Vert_V (\Vert u \Vert_V^{-1} u), \tag{12}$
with
$\Vert \Vert u \Vert_V^{-1} u \Vert_V = \Vert u \Vert_V^{-1} \Vert u \Vert_V = 1; \tag{13}$
then by (11)-(13),
$\Vert u \Vert_V^{-1} \Vert T(u) \Vert_W = \Vert \Vert u \Vert_V^{-1} T(u) \Vert_W = \Vert T(\Vert u \Vert_V^{-1} u) \Vert_W <$
$\dfrac{\epsilon}{r \delta} \Vert \Vert u \Vert_V^{-1} u \Vert_V = \dfrac{\epsilon}{r \delta} \Vert u \Vert_V^{-1} \Vert u \Vert_V, \tag{14}$
whence
$ \Vert T(u) \Vert_W < \dfrac{\epsilon}{r \delta} \Vert u \Vert_V; \tag{15}$
we can include the case $u = 0$ by replacing the "$<$" sign with "$\le$":
$ \Vert T(u) \Vert_W \le \dfrac{\epsilon}{r \delta} \Vert u \Vert_V, \tag{16}$
holding for all $u \in V$.  (16) binds for all $r$, $0 < r < 1$; thus letting
$r \to 1^-$ we may affirm that 
$ \Vert T(u) \Vert_W \le \dfrac{\epsilon}{\delta} \Vert u \Vert_V \tag{17}$
for any $u \in V$; thus $T$ is a bounded operator, with bound no greater than
$\epsilon / \delta$.
A: Your prove is correct. An alternative prove for both implications can be found on this course page 14 
