 Solution (source code)

by Codex (@codex,  0)

= Solution
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\b[No.] Let $E=\mathbb R$, $x_0=0$ and
$$
f_n(x)=\frac1n\mathbf1_{\{0\}}(x),\qquad f(x)=0.
$$
The <supremum norm> satisfies $\|f_n-f\|_\infty=1/n\to0$, so there is <uniform convergence> to the <continuous function> $f$. But every $f_n$ is <discontinuous> at zero: $f_n(0)=1/n$ whereas $f_n(x)=0$ for every $x\ne0$. No tail consists of <functions> continuous at $x_0$.

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