Is the strong topology on $B(H)$ first countable? Let $H$ be a separable Hilbert space and consider the strong operator topology (SOT) on $B(H)$, that is the topology on $B(H)$ generated by the seminorms
$$p_x: u \mapsto \Vert ux \Vert, \quad x\in H$$
Is this topology first countable?
Attempt: It suffices to find a countable neighborhood basis of $0$. Let $(x_n)_n$ be a dense sequence of $H$. Then I believe the topology generated by $$u \mapsto \Vert u x_n \Vert, \quad n \geq 1$$
agrees with SOT. Hence, the collection
$$\left\{\bigcap_{i=1}^k p_{x_{n_i}}^{-1}([0, \epsilon[): k \geq 1; n_1, \dots, n_k \geq 1, \epsilon \in \Bbb{Q}\cap(0, \infty) \right\}$$
is a countable basis of $0$ of $SOT$.
Is the above proof outline correct?
EDIT: This question was accidentally closed by a misunderstanding. Please read the comments and reopen the question. Thanks!
 A: Theorem.  If  $H$ is an infinite dimensional Hilbert space  then the strong topology on $B(H)$ is not first countable.
Proof. Suppose by way contradiction that $B(H)$ is first countable and hence we may pick  a countable neighborhood basis
of the zero operator, say $\{U_n\}_{n\in {\mathbb N}}$.  For each $n$, choose a neighborhood $V_n$ of  zero of the form
$$
  V_n=\{T\in B(H): \Vert T(x^n_i)\Vert <\varepsilon ^n_i,  \ i=1, 2, \ldots , k_n\},
  $$
with $V_n\subseteq U_n$,
where the $x^n_i$ are vectors in $H$ and the $\varepsilon ^n_i$ are positive real numbers.
Let $X$ denote the set formed by all of the $x^n_i$ showing up in the above descriptions of all of the $V_n$,    clearly
a  countable set.
Since $H$ is infinite dimensional, its (Hamel) dimension is uncountable (regardless of $H$ being separable or not), so
the linear span of $X$  (no closure) is a proper subset of $H$.
Fixing a vector $y$ in $H\setminus \text{span}(X)$, consider the neighborhood of zero given by
$$
  V=\{T\in B(H): \Vert T(y)\Vert <1\}.
  $$
Given that   $\{U_n\}_{n\in {\mathbb N}}$ is a  neighborhood basis for zero, there ought to be some $n$ such that $U_n\subseteq V$, and
hence also that $V_n\subseteq V$.
Interpreting the last inclusion we see that for every bounded linear operator $T$ on $H$ one has that
$$
  \Vert T(x^n_i)\Vert <\varepsilon ^n_i,\ \forall i=1, 2, \ldots , k_n \quad \Rightarrow \quad\Vert T(y)\Vert <1.
  $$
Observing that  $y$ is not in the span of
$\{x^n_i:   i=1, 2, \ldots , k_n\}$, (it is not  even in $\text{span}(X)$), we may easily construct a
bounded operator $T$ such that $T(x^n_i)=0$, for all $i$, while $\Vert T(y)\Vert =1$, and this operator contradicts the above  implication.
QED.

Remarks:  Assuming WLOG that we choose the vector $y$ in the above proof with $\Vert y\Vert = 1$, the operator $T$ with smallest norm satisfying the above conditions has norm equal to
$$
  {1 \over \text{dist}\big(y, \text{span}\{x^n_i:   i=1, 2, \ldots , k_n\}\big)}.
  $$
Therefore this will explode with $n$ in case
the span of
$\{x^n_i:   i=1, 2, \ldots , k_n\}$ gets too close to $y$.
This fact is behind  the reason why the strong topology is not first countable on $B(H)$ despite being
metrizable, and hence also
first countable, when  restricted to a bounded subset of $B(H)$, such as the set $U(H)$ of
all unitary operators (cf. The strong topology on $U(\mathcal H)$ is metrisable).
