= Solution
The <Baire category theorem> states that a nonempty complete <metric space> cannot be a countable union of closed <nowhere dense sets>. The <Uniform boundedness principle> says that if $V$ is a <Banach space>, $W$ a <normed vector space>, and $\mathcal T\subseteq B(V,W)$ is <pointwise> bounded, meaning $\sup_{T\in\mathcal T}\|Tx\|<\infty$ for every $x\in V$, then $\sup_{T\in\mathcal T}\|T\|<\infty$.
To prove it, let $E_n=\{x:\sup_{T\in\mathcal T}\|Tx\|\leq n\}$. Each set is closed by <continuity> of the individual operators, and their union is $V$. The <Baire category theorem> gives a ball $B(x_0,r)$ inside some $E_N$. For $\|h\|<r$, both $x_0+h$ and $x_0$ are in $E_N$, so $\|Th\|\leq\|T(x_0+h)\|+\|Tx_0\|\leq2N$ for every $T$. Scaling gives $\|T\|\leq2N/r$ uniformly. The zero-space and empty-family cases are immediate.
For the first family on a <Hilbert space>, fix $x$. The functionals $y\mapsto\langle y,Tx\rangle$ have <operator norm> $\|Tx\|$ by the <Cauchy-Schwarz inequality> and equality on the direction of $Tx$. Their values are <pointwise> bounded by the hypothesis, after taking complex conjugates if necessary. Apply the <Uniform boundedness principle> to obtain $\sup_{T\in\mathcal E}\|Tx\|<\infty$. Apply it a second time to the operator family itself to conclude \b[$\sup_{T\in\mathcal E}\|T\|<\infty$].
For the diagonal hypothesis, the answer depends on the scalar <field>. Over $\mathbb C$, let $q_T(x)=\langle Tx,x\rangle$, using the convention that the <inner product> is linear in its first variable. The complex <polarization identity> gives
$$
4\langle Tx,y\rangle=q_T(x+y)-q_T(x-y)+i q_T(x+iy)-i q_T(x-iy).
$$
The assumed bounds at these four fixed vectors make every matrix coefficient uniformly bounded, so the preceding argument applies. Thus \b[the complex-space answer is yes], even without assuming self-adjointness.
Over $\mathbb R$, take $X=\mathbb R^2$ and
$$
J=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad\mathcal F=\{nJ:n\geq1\}.
$$
Every diagonal value $\langle nJx,x\rangle$ is zero, whereas $\|nJ\|=n$. Thus \b[the real-space answer is no in general]. Real diagonal values do not detect the skew-symmetric part of an operator. This is <diagonal boundedness of operator families over real and complex Hilbert spaces>.
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