Consider an infinite dimensional Banach space $(X,||\cdot||)$. Let $\mathcal{P}$ be the family of all equivalent norms in $X$. That is $p\in \mathcal{P}$ iff $p$ is equivalent to $||\cdot||$, i.e. there exists two constants $c_1,c_2>0$ such that $c_1||x||\leq p(x)\leq c_2||x||$ for all $x\in X$.
$\mathcal{P}$ is a metric space endowed with the following metric $\rho$:
$$\rho(p,q)=\sup_{||x||\leq 1} \lbrace |p(x)-q(x)|\rbrace$$
$\mathcal{P}$ is a Baire space, seen as an open subset of the space of all continuous semi-norms in $(X,||\cdot||)$ with the same metric $\rho$. This last space is a complete metric space, so by the Baire category theorem it is a Baire space.
Is $\mathcal{P}$ a locally compact space? - We can't use Riesz Lemma, as this space is not a linear space, to argue against the locally compactness.
No, here is a counterexample. Take $X=L^1(\mathbb{R})$, $\|.\|_1$ the usual $L^1$ norm.
By contradiction, assume $\mathcal{P}$ to be locally compact. Then there exists a compact neighbourhood $K$ of $\|.\|$ which is compact. Let $\epsilon>0$ be such that $B(\|.\|_1,2\epsilon)$ is a subset of $K$.
Define for $f \in X$, $$p_n(f)=\int_{\mathbb{R}} |f(t)|\left(1+\epsilon e^{-(t-n)^2}\right)dt.$$
These norms are equivalent to $\|.\|_1$ (with $c_1=1$ and $c_2=1+\epsilon$).
Clearly, $\rho(p_n,\|.\|_1)\leq \epsilon$ and in fact there is equality (by considering a sequence $(f_k)_k$ of approximation of unity around $t=n$).
Since $p_n \in K$ for every $n$, one can extract a subsequence converging. However, no extracted sequence of $(p_n)_n$ is a Cauchy sequence: for every $n$, $$\lim_{m\to +\infty} \rho(p_n,p_m)\geq \epsilon,$$ again by considering a sequence $(f_k)_k$ of approximation of unity around $t=n$. This yields a contradiction.