Iosif Pinelis neatly answered the question. I just want to add another viewpoint. The Minkowski functional (or gauge) $p_B(x)=\inf\{t>0: x\in tB\}$ of an absolutely convex absorbing set is a semi-norm which almost gives back $B$ since $\{p_B<1\}\subseteq B\subseteq \{p_B\le 1\}$.

Since two semi-norms $p_k$ on $V$ with unit balls $A_k=\{p_k\le 1\}$ satisfy $p_1\le p_2$ if and only if $A_2\subseteq A_1$, it is immediate that the Minkowski functional of the convex (=absolutely convex) hull of $B_1\cup B_2$ (where $B_k$ are the unit balls of $\|\cdot\|_k$) is the infimum of $\|\cdot\|_1$ and $\|\cdot\|_2$ in the partially ordered set of semi-norms on $V$. One has thus to check that $\|\cdot\|_+$ is the infimum of $\|\cdot\|_1$ and $\|\cdot\|_2$: On the one hand, $\|v\|_+\le\|v\|_k$ since $v=0+v=v+0$ and, on the other hand, if $p$ is a seminorm with $p\le\|\cdot\|_k$ the triangle inequality yields $p\le\|\cdot\|_+$.

I doubt, by the way, that $\|\cdot\|_+$ is always a proper norm (it is of course a semi-norm): Any two separable Banach spaces have the same algebraic dimension (namely, the cardinality of $\mathbb R$, see, e.g., https://math.stackexchange.com/questions/1899365/hamel-dimension-of-infinite-dimensional-separable-banach-space, however, for a simple example, consider $\ell^2\subseteq c_0$, so that dim$(\ell^2)\le$dim$(c_0)$, but on the other hand, $c_0\to\ell^2$, $(x_n)_{n\in\mathbb N}\mapsto (x_n/n)_{n\in\mathbb N}$ is a linear injection which implies dim$(c_0)\le$dim$(\ell^2)$, talking about Hamel dimension requires the axiom of choice). We can thus find two incomparable complete norms $\|\cdot\|_k$ on a vector space $V$. If $\|\cdot\|_+$ were a proper norm, the identical map $i:(V,\|\cdot\|_1)\to (V,\|\cdot\|_2)$ would have closed graph because it is continuous if both spaces are endowed with the coarser Hausdorff topology generated by $\|\cdot\|_+$ (if one prefers an elementary argument, for $x_n\to x$ in $(V;\|\cdot\|_1)$ with $i(x_n)\to y$ in $(V,\|\cdot\|_2)$ we have $i(x)=y$ because $\|\cdot\|_k$-convergence implies $\|\cdot\|_+$-convergence and $\|\cdot\|_+$-limits are unique). The closed graph theorem implies that $i$ is continuous so that $\|\cdot\|_2\le c\|\cdot\|_1$ for some constant $c$, contradicting the incomparability of the norms.

When talking about closed unit balls one should thus state precisely the topology in which the ball is closed.

Iosif Pinelis neatly answered the question. I just want to add another viewpoint. The Minkowski functional (or gauge) $p_B(x)=\inf\{t>0: x\in tB\}$ of an absolutely convex absorbing set is a semi-norm which almost gives back $B$ since $\{p_B<1\}\subseteq B\subseteq \{p_B\le 1\}$.

Since two semi-norms $p_k$ on $V$ with unit balls $A_k=\{p_k\le 1\}$ satisfy $p_1\le p_2$ if and only if $A_2\subseteq A_1$, it is immediate that the Minkowski functional of the convex (=absolutely convex) hull of $B_1\cup B_2$ (where $B_k$ are the unit balls of $\|\cdot\|_k$) is the infimum of $\|\cdot\|_1$ and $\|\cdot\|_2$ in the partially ordered set of semi-norms on $V$. One has thus to check that $\|\cdot\|_+$ is the infimum of $\|\cdot\|_1$ and $\|\cdot\|_2$: On the one hand, $\|v\|_+\le\|v\|_k$ since $v=0+v=v+0$ and, on the other hand, if $p$ is a seminorm with $p\le\|\cdot\|_k$ the triangle inequality yields $p\le\|\cdot\|_+$.

I doubt, by the way, that $\|\cdot\|_+$ is always a proper norm (it is of course a semi-norm): Any two separable Banach spaces have the same algebraic dimension (namely, the cardinality of $\mathbb R$, see, e.g., https://math.stackexchange.com/questions/1899365/hamel-dimension-of-infinite-dimensional-separable-banach-space, however, for a simple example, consider $\ell^2\subseteq c_0$, so that dim$(\ell^2)\le$dim$(c_0)$, but on the other hand, $c_0\to\ell^2$, $(x_n)_{n\in\mathbb N}\mapsto (x_n/n)_{n\in\mathbb N}$ is a linear injection which implies dim$(c_0)\le$dim$(\ell^2)$, talking about Hamel dimension requires the axiom of choice). We can thus find two incomparable complete norms $\|\cdot\|_k$ on a vector space $V$ (e.g., $V=\ell^2$ with the usual norm $\|\cdot\|_2$ and $\|v\|_1=\|T(v)\|_{c_0}$ for a linear bijection $T:\ell_2\to c_0$ which can't be continuous since $\ell_2$ and $c_0$ are not isomorphic as Banach spaces). If $\|\cdot\|_+$ were a proper norm, the identical map $i:(V,\|\cdot\|_1)\to (V,\|\cdot\|_2)$ would have closed graph because it is continuous if both spaces are endowed with the coarser Hausdorff topology generated by $\|\cdot\|_+$ (if one prefers an elementary argument, for $x_n\to x$ in $(V;\|\cdot\|_1)$ with $i(x_n)\to y$ in $(V,\|\cdot\|_2)$ we have $i(x)=y$ because $\|\cdot\|_k$-convergence implies $\|\cdot\|_+$-convergence and $\|\cdot\|_+$-limits are unique). The closed graph theorem implies that $i$ is continuous so that $\|\cdot\|_2\le c\|\cdot\|_1$ for some constant $c$, contradicting the incomparability of the norms.

When talking about closed unit balls one should thus state precisely the topology in which the ball is closed.

Iosif Pinelis neatly answered the question. I just want to add another viewpoint. The Minkowski functional (or gauge) $p_B(x)=\inf\{t>0: x\in tB\}$ of an absolutely convex absorbing set is a semi-norm which almost gives back $B$ since $\{p_B<1\}\subseteq B\subseteq \{p_B\le 1\}$.

Since two semi-norms $p_k$ on $V$ with unit balls $A_k=\{p_k\le 1\}$ satisfy $p_1\le p_2$ if and only if $A_2\subseteq A_1$, it is immediate that the Minkowski functional of the convex (=absolutely convex) hull of $B_1\cup B_2$ (where $B_k$ are the unit balls of $\|\cdot\|_k$) is the infimum of $\|\cdot\|_1$ and $\|\cdot\|_2$ in the partially ordered set of semi-norms on $V$. One has thus to check that $\|\cdot\|_+$ is the infimum of $\|\cdot\|_1$ and $\|\cdot\|_2$: On the one hand, $\|v\|_+\le\|v\|_k$ since $v=0+v=v+0$ and, on the other hand, if $p$ is a seminorm with $p\le\|\cdot\|_k$ the triangle inequality yields $p\le\|\cdot\|_+$.

I doubt, by the way, that $\|\cdot\|_+$ is always a proper norm (it is of course a semi-norm): Any two separable Banach spaces have the same algebraic dimension so that the axiom of choice allows to find two incomparable complete norms $\|\cdot\|_k$ on a vector space $V$. If $\|\cdot\|_+$ were a proper norm, the identical map $i:(V,\|\cdot\|_1)\to (V,\|\cdot\|_2)$ would have closed graph implying its continuity so that $\|\cdot\|_2\le c\|\cdot\|_1$ for some constant $c$.

When talking about closed unit balls one should thus state precisely the topology in which the ball is closed.

Iosif Pinelis neatly answered the question. I just want to add another viewpoint. The Minkowski functional (or gauge) $p_B(x)=\inf\{t>0: x\in tB\}$ of an absolutely convex absorbing set is a semi-norm which almost gives back $B$ since $\{p_B<1\}\subseteq B\subseteq \{p_B\le 1\}$.

Since two semi-norms $p_k$ on $V$ with unit balls $A_k=\{p_k\le 1\}$ satisfy $p_1\le p_2$ if and only if $A_2\subseteq A_1$, it is immediate that the Minkowski functional of the convex (=absolutely convex) hull of $B_1\cup B_2$ (where $B_k$ are the unit balls of $\|\cdot\|_k$) is the infimum of $\|\cdot\|_1$ and $\|\cdot\|_2$ in the partially ordered set of semi-norms on $V$. One has thus to check that $\|\cdot\|_+$ is the infimum of $\|\cdot\|_1$ and $\|\cdot\|_2$: On the one hand, $\|v\|_+\le\|v\|_k$ since $v=0+v=v+0$ and, on the other hand, if $p$ is a seminorm with $p\le\|\cdot\|_k$ the triangle inequality yields $p\le\|\cdot\|_+$.

I doubt, by the way, that $\|\cdot\|_+$ is always a proper norm (it is of course a semi-norm): Any two separable Banach spaces have the same algebraic dimension (namely, the cardinality of $\mathbb R$, see, e.g., https://math.stackexchange.com/questions/1899365/hamel-dimension-of-infinite-dimensional-separable-banach-space, however, for a simple example, consider $\ell^2\subseteq c_0$, so that dim$(\ell^2)\le$dim$(c_0)$, but on the other hand, $c_0\to\ell^2$, $(x_n)_{n\in\mathbb N}\mapsto (x_n/n)_{n\in\mathbb N}$ is a linear injection which implies dim$(c_0)\le$dim$(\ell^2)$, talking about Hamel dimension requires the axiom of choice). We can thus find two incomparable complete norms $\|\cdot\|_k$ on a vector space $V$. If $\|\cdot\|_+$ were a proper norm, the identical map $i:(V,\|\cdot\|_1)\to (V,\|\cdot\|_2)$ would have closed graph because it is continuous if both spaces are endowed with the coarser Hausdorff topology generated by $\|\cdot\|_+$ (if one prefers an elementary argument, for $x_n\to x$ in $(V;\|\cdot\|_1)$ with $i(x_n)\to y$ in $(V,\|\cdot\|_2)$ we have $i(x)=y$ because $\|\cdot\|_k$-convergence implies $\|\cdot\|_+$-convergence and $\|\cdot\|_+$-limits are unique). The closed graph theorem implies that $i$ is continuous so that $\|\cdot\|_2\le c\|\cdot\|_1$ for some constant $c$, contradicting the incomparability of the norms.

When talking about closed unit balls one should thus state precisely the topology in which the ball is closed.