= Solution
First prove the <Baire category theorem> in its complete-metric form. Let $X$ be a nonempty <complete metric space>, let $G_n$ be open <dense sets>, and let $V$ be any nonempty <open set>. Choose a closed ball $\overline B(x_1,r_1)\subset V\cap G_1$ with $0<r_1<1/2$. Inductively, density and openness permit the choice
$$
\overline B(x_{n+1},r_{n+1})\subset B(x_n,r_n)\cap G_{n+1},\qquad 0<r_{n+1}<2^{-(n+1)}.
$$
For $m>n$, $x_m$ belongs to the $n$th closed ball, so $d(x_m,x_n)\le r_n$. The centers form a <Cauchy sequence> and converge to some $x\in X$ by completeness. Each closed ball contains all later centers and therefore contains their limit. Hence $x\in V\cap\bigcap_nG_n$. Since this holds for every $V$, \b[every countable intersection of open dense sets is dense]. Equivalently, a nonempty <complete metric space> cannot be a countable union of closed sets with empty interior; this is the form needed below.
For the <Banach-Steinhaus theorem>, let $X$ be a <Banach space>, $Y$ a <normed vector space>, and $\mathcal T$ a family of <bounded linear operators> from $X$ to $Y$ such that $\sup_{T\in\mathcal T}\|Tx\|<\infty$ for every $x$. Define
$$
E_m=\{x\in X:\|Tx\|\le m\text{ for every }T\in\mathcal T\},\qquad m=1,2,\ldots.
$$
Each $E_m$ is closed, as an intersection of inverse images of closed balls under continuous maps, and pointwise boundedness gives $X=\bigcup_mE_m$. The <Baire category theorem> gives some $E_m$ with nonempty interior. Choose $B(x_0,r)\subset E_m$. If $\|h\|\le r/2$, both $x_0+h$ and $x_0$ lie in $E_m$, so $\|Th\|\le2m$ for every $T$. Applying this to $h=(r/2)u$ with $\|u\|\le1$ proves
$$
\boxed{\sup_{T\in\mathcal T}\|T\|\le\frac{4m}{r}<\infty.}
$$
This proves the <Uniform boundedness principle>; completeness is required for the domain, but not for $Y$.
Now fix the prescribed point $\theta_0$ on the <unit circle>, parametrized by angles modulo $2\pi$. On the real <Banach space> $C(\mathbb T;\mathbb R)$ with the <supremum norm>, consider the <continuous linear functionals> $T_Nf=S_Nf(\theta_0)$, where $S_N$ is the symmetric <Fourier partial sum>. Summing the finite <geometric series> gives its <Dirichlet kernel> representation
$$
T_Nf=\frac1{2\pi}\int_{-\pi}^{\pi}f(\theta_0-t)D_N(t)\,dt,\qquad D_N(t)=\sum_{j=-N}^Ne^{ijt}=\frac{\sin((N+1/2)t)}{\sin(t/2)}.
$$
The continuous extension at $t=0$ is $2N+1$. The kernel bound gives $\|T_N\|\le\Lambda_N=(2\pi)^{-1}\int|D_N|$. Equality holds: the continuous periodic functions defined by
$$
f_\delta(\theta_0-t)=\frac{D_N(t)}{\sqrt{D_N(t)^2+\delta^2}}
$$
have <supremum norm> at most one, and <dominated convergence> gives $T_Nf_\delta\to\Lambda_N$ as $\delta\downarrow0$.
These norms are unbounded. Indeed, evenness and $\sin(t/2)\le t/2$ for $0<t\le\pi$ give
$$
\Lambda_N\ge\frac2\pi\int_0^{(N+1/2)\pi}\frac{|\sin u|}{u}\,du\ge\frac4{\pi^2}\sum_{j=1}^N\frac1j.
$$
For the second inequality, integrate over $[(j-1)\pi,j\pi]$, where $1/u\ge1/(j\pi)$ and the integral of $|\sin u|$ is two. The <harmonic series> diverges, proving the <Dirichlet kernel harmonic lower bound>. If every continuous $f$ had bounded partial sums at $\theta_0$, the <Banach-Steinhaus theorem> would bound all the operator norms, a contradiction. Consequently \b[there exists a continuous real function $f$ with $\sup_N|S_Nf(\theta_0)|=\infty$], and its <Fourier series> diverges at the prescribed point. The function depends on the chosen point; the argument requires no claim of simultaneous divergence everywhere.
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