= Solution
Work over $\mathbb F=\mathbb R$ or $\mathbb C$. A <Fréchet space> is a Hausdorff, complete, metrizable <locally convex space>. Its <topology> is generated by a countable separating family of <seminorms> $(p_n)_{n\ge1}$. A neighbourhood base at zero consists of finite intersections
$$
\{x:p_{n_1}(x)<\varepsilon_1,\ldots,p_{n_k}(x)<\varepsilon_k\}.
$$
Replacing $p_n$ by $\max_{j\le n}p_j$ makes the family increasing. One compatible translation-invariant <metric> is
$$
d(x,y)=\sum_{n\ge1}2^{-n}\min\{1,p_n(x-y)\}.
$$
Completeness means that a sequence Cauchy for all these <seminorms> converges in the space. The separating condition ensures that $d(x,y)=0$ only when $x=y$.
Let $E'$ denote the <continuous dual space> for the original <topology> $\tau$. The <weak topology> $\sigma(E,E')$ is the coarsest <topology> making all $f\in E'$ continuous. Its basic zero-neighbourhoods are
$$
\{x:|f_1(x)|<\varepsilon_1,\ldots,|f_k(x)|<\varepsilon_k\},
\qquad f_j\in E'.
$$
Equivalently, it is generated by the <seminorms> $x\mapsto|f(x)|$.
To prove the Hausdorff property, fix $z\ne0$. Some continuous <seminorm> $p$ has $p(z)>0$. On the one-dimensional span of $z$, define $f_0(az)=a\,p(z)$. Then $|f_0(az)|=p(az)$. The real or complex <Hahn-Banach theorem> extends this to $f:E\to\mathbb F$ with $|f(x)|\le p(x)$. Thus $f$ is originally continuous and $f(z)\ne0$. This proves <continuous-dual point separation>. For distinct $x,y$, inverse images under such an $f$ of disjoint small scalar neighbourhoods of $f(x)$ and $f(y)$ are disjoint weak neighbourhoods. Hence \b[the <weak topology> is Hausdorff].
If a <linear functional> is originally continuous, it belongs to $E'$ and is weakly continuous by definition. Conversely, $\sigma(E,E')$ is coarser than $\tau$, so every weakly continuous scalar function is originally continuous. In particular,
$$
\boxed{(E,\sigma(E,E'))'=E'.}
$$
This special case of the <continuous dual of a weak topology> needs no completeness argument.
For the <linear map> $T:E_1\to E_2$, original continuity implies that $g\circ T\in E_1'$ whenever $g\in E_2'$. Each such composition is weakly continuous on $E_1$, so the defining property of the target <weak topology> makes $T$ weak-to-weak continuous.
Conversely, suppose $T$ is weak-to-weak continuous. Every $g\circ T$, $g\in E_2'$, is then weakly continuous, and hence continuous for $\tau_1$. We show that the <graph of a linear operator> $\Gamma(T)$ is closed in the original product <topology>. If $x_n\to x$ in $E_1$ and $Tx_n\to y$ in $E_2$, then
$$
g(y)=\lim_n g(Tx_n)=\lim_n(g\circ T)(x_n)=g(Tx)
\qquad(g\in E_2').
$$
By <continuous-dual point separation> on $E_2$, this forces $y=Tx$. The product is metrizable, so sequential closedness is closedness. The <closed graph theorem for Fréchet spaces> now gives original continuity of $T$.
The relevant version of that theorem does use completeness of both spaces: a closed graph is a <Fréchet space>, its first-coordinate projection is a continuous linear bijection onto $E_1$, and the <open mapping theorem for Fréchet spaces> gives a continuous inverse. Composing with the second-coordinate projection gives $T$. Thus
$$
\boxed{T\text{ is originally continuous}\iff T\text{ is weak-to-weak continuous}.}
$$
Equality of continuous scalar functionals does not imply equality of the two <topologies>; the <closed graph theorem for Fréchet spaces> is the extra step that makes this conclusion valid.
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