At the $47$-th paragraph on the text Topology by James Munkres is asked to prove the following result
Theorem
If $(X,\cal T)$ is a hemicompact and $\underline{\text{hausdorff}}$ space then any sequence $(f_n)_{n\in\omega}$ of functions from $X$ to $\Bbb R^n$ with $n\in\omega_+$ has a compactly convergent subsequence whether it is equicontinuous and pointwise bounded.
So I proved the above theorem as to follow.
Prof. So if $(f_n)_{n\in\omega}$ is equicontinuous then by definition for any $x\in X$ the set $$ \cal F_x:=\big\{f_n(x):n\in\omega\big\} $$ is bounded so that it has compact clausure and thus by Ascoli's theorem the set $$ \cal F:=\{f_n:n\in\omega\} $$ is contained in a compact subspace $\cal K$ of $\mathcal C(X,\Bbb R^n)$ equipped with compact convergence topology $\cal T_k$. After all, by emicompactness we know that $(Y^X,\cal T_k)$ is metrizable so that $\cal K$ is sequentially compact because any subspace of a metrizable space is metrizable and because for metric space compactness and sequential compactness are equivalent. So by sequentially compactness of $\cal K$ we conclude that $(f_n)_{n\in\omega}$ has a subsequence converging to any element $f$ of $\cal K$ or rather of $\mathcal C(X,\Bbb R^n)$.
So it seem I to me I well proved the theorem but it seem to me even I did not use hausdorfness so that I thought to put a specific question where I ask clarification about hausdorfness: so did I well prove the theorem? is actually hausdorfness necessary and did I use it? Could someone help me, please?
N.B.
Really Munkres do not requires that $X$ is hemicompact with respect $\cal T$ but $\sigma$-compact: however, $\sigma$-compactness defined by Munkres is not the classical one but it is a special case of hemicompactness so that I think that this is actually the reason for which hausdorfness is necessary for Munkres but obviously this is only my opinion; anyway, click here if you like to know Munkres $\sigma$-compactness.
